#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head
int _,n;
namespace linear_seq {
const int N=10010;
ll res[N],base[N],_c[N],_md[N];
vector<int> Md;
void mul(ll *a,ll *b,int k) {
rep(i,0,k+k) _c[i]=0;
rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
for (int i=k+k-1;i>=k;i--) if (_c[i])
rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
rep(i,0,k) a[i]=_c[i];
}
int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
// printf("%d\n",SZ(b));
ll ans=0,pnt=0;
int k=SZ(a);
assert(SZ(a)==SZ(b));
rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
Md.clear();
rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
rep(i,0,k) res[i]=base[i]=0;
res[0]=1;
while ((1ll<<pnt)<=n) pnt++;
for (int p=pnt;p>=0;p--) {
mul(res,res,k);
if ((n>>p)&1) {
for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
}
}
rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
if (ans<0) ans+=mod;
return ans;
}
VI BM(VI s) {
VI C(1,1),B(1,1);
int L=0,m=1,b=1;
rep(n,0,SZ(s)) {
ll d=0;
rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
VI T=C;
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(0);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; b=d; m=1;
} else {
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(0);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
return C;
}
int gao(VI a,ll n) {
VI c=BM(a);
c.erase(c.begin());
rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}
};
int main() {
while (~scanf("%d",&n)) {
vector<int>v;
v.push_back(1);
v.push_back(2);
v.push_back(4);
v.push_back(7);
v.push_back(13);
v.push_back(24);
//VI{1,2,4,7,13,24}
printf("%d\n",linear_seq::gao(v,n-1));
}
}
以下既可以处理模数为素数的情况，也可以处理模数不是素数的情况，需要代入初始数据
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <unordered_map>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef vector<long long> VI;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const ll mod=1e9;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define SZ(x) ((long long)(x).size())
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
ll powmod(ll a,ll b)
{
ll res=1ll;
while(b)
{
if(b&1) res=res*a%mod;
a=a*a%mod,b>>=1;
}
return res;
}
namespace linear_seq {
const int N=10010;
using int64 = long long;
using vec = std::vector<int64>;
ll res[N],base[N],_c[N],_md[N];
vector<int> Md;
void mul(ll *a,ll *b,int k) {
FOR(i,0,k+k-1) _c[i]=0;
FOR(i,0,k-1) if (a[i]) FOR(j,0,k-1) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
for (int i=k+k-1;i>=k;i--) if (_c[i])
FOR(j,0,SZ(Md)-1) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
FOR(i,0,k-1) a[i]=_c[i];
}
int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
// printf("%d\n",SZ(b));
ll ans=0,pnt=0;
int k=SZ(a);
assert(SZ(a)==SZ(b));
FOR(i,0,k-1) _md[k-1-i]=-a[i];_md[k]=1;
Md.clear();
FOR(i,0,k-1) if (_md[i]!=0) Md.push_back(i);
FOR(i,0,k-1) res[i]=base[i]=0;
res[0]=1;
while ((1ll<<pnt)<=n) pnt++;
for (int p=pnt;p>=0;p--) {
mul(res,res,k);
if ((n>>p)&1) {
for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
FOR(j,0,SZ(Md)-1) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
}
}
FOR(i,0,k-1) ans=(ans+res[i]*b[i])%mod;
if (ans<0) ans+=mod;
return ans;
}
VI BM(VI s) {
VI C(1,1),B(1,1);
int L=0,m=1,b=1;
FOR(n,0,SZ(s)-1) {
ll d=0;
FOR(i,0,L) d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
VI T=C;
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(0);
FOR(i,0,SZ(B)-1) C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; b=d; m=1;
} else {
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(0);
FOR(i,0,SZ(B)-1) C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
return C;
}
static void extand(vec &a, size_t d, int64 value = 0) {
if (d <= a.size()) return;
a.resize(d, value);
}
static void exgcd(int64 a, int64 b, int64 &g, int64 &x, int64 &y) {
if (!b) x = 1, y = 0, g = a;
else {
exgcd(b, a % b, g, y, x);
y -= x * (a / b);
}
}
static int64 crt(const vec &c, const vec &m) {
int n = c.size();
int64 M = 1, ans = 0;
for (int i = 0; i < n; ++i) M *= m[i];
for (int i = 0; i < n; ++i) {
int64 x, y, g, tm = M / m[i];
exgcd(tm, m[i], g, x, y);
ans = (ans + tm * x * c[i] % M) % M;
}
return (ans + M) % M;
}
static vec ReedsSloane(const vec &s, int64 mod) {
auto inverse = [](int64 a, int64 m) {
int64 d, x, y;
exgcd(a, m, d, x, y);
return d == 1 ? (x % m + m) % m : -1;
};
auto L = [](const vec &a, const vec &b) {
int da = (a.size() > 1 || (a.size() == 1 && a[0])) ? a.size() - 1 : -1000;
int db = (b.size() > 1 || (b.size() == 1 && b[0])) ? b.size() - 1 : -1000;
return std::max(da, db + 1);
};
auto prime_power = [&](const vec &s, int64 mod, int64 p, int64 e) {
// linear feedback shift register mod p^e, p is prime
std::vector<vec> a(e), b(e), an(e), bn(e), ao(e), bo(e);
vec t(e), u(e), r(e), to(e, 1), uo(e), pw(e + 1);;
pw[0] = 1;
for (int i = pw[0] = 1; i <= e; ++i) pw[i] = pw[i - 1] * p;
for (int64 i = 0; i < e; ++i) {
a[i] = {pw[i]}, an[i] = {pw[i]};
b[i] = {0}, bn[i] = {s[0] * pw[i] % mod};
t[i] = s[0] * pw[i] % mod;
if (t[i] == 0) {
t[i] = 1, u[i] = e;
} else {
for (u[i] = 0; t[i] % p == 0; t[i] /= p, ++u[i]);
}
}
for (size_t k = 1; k < s.size(); ++k) {
for (int g = 0; g < e; ++g) {
if (L(an[g], bn[g]) > L(a[g], b[g])) {
ao[g] = a[e - 1 - u[g]];
bo[g] = b[e - 1 - u[g]];
to[g] = t[e - 1 - u[g]];
uo[g] = u[e - 1 - u[g]];
r[g] = k - 1;
}
}
a = an, b = bn;
for (int o = 0; o < e; ++o) {
int64 d = 0;
for (size_t i = 0; i < a[o].size() && i <= k; ++i) {
d = (d + a[o][i] * s[k - i]) % mod;
}
if (d == 0) {
t[o] = 1, u[o] = e;
} else {
for (u[o] = 0, t[o] = d; t[o] % p == 0; t[o] /= p, ++u[o]);
int g = e - 1 - u[o];
if (L(a[g], b[g]) == 0) {
extand(bn[o], k + 1);
bn[o][k] = (bn[o][k] + d) % mod;
} else {
int64 coef = t[o] * inverse(to[g], mod) % mod * pw[u[o] - uo[g]] % mod;
int m = k - r[g];
extand(an[o], ao[g].size() + m);
extand(bn[o], bo[g].size() + m);
for (size_t i = 0; i < ao[g].size(); ++i) {
an[o][i + m] -= coef * ao[g][i] % mod;
if (an[o][i + m] < 0) an[o][i + m] += mod;
}
while (an[o].size() && an[o].back() == 0) an[o].pop_back();
for (size_t i = 0; i < bo[g].size(); ++i) {
bn[o][i + m] -= coef * bo[g][i] % mod;
if (bn[o][i + m] < 0) bn[o][i + m] -= mod;
}
while (bn[o].size() && bn[o].back() == 0) bn[o].pop_back();
}
}
}
}
return std::make_pair(an[0], bn[0]);
};
std::vector<std::tuple<int64, int64, int>> fac;
for (int64 i = 2; i * i <= mod; ++i)
if (mod % i == 0) {
int64 cnt = 0, pw = 1;
while (mod % i == 0) mod /= i, ++cnt, pw *= i;
fac.emplace_back(pw, i, cnt);
}
if (mod > 1) fac.emplace_back(mod, mod, 1);
std::vector<vec> as;
size_t n = 0;
for (auto &&x: fac) {
int64 mod, p, e;
vec a, b;
std::tie(mod, p, e) = x;
auto ss = s;
for (auto &&x: ss) x %= mod;
std::tie(a, b) = prime_power(ss, mod, p, e);
as.emplace_back(a);
n = std::max(n, a.size());
}
vec a(n), c(as.size()), m(as.size());
for (size_t i = 0; i < n; ++i) {
for (size_t j = 0; j < as.size(); ++j) {
m[j] = std::get<0>(fac[j]);
c[j] = i < as[j].size() ? as[j][i] : 0;
}
a[i] = crt(c, m);
}
return a;
}
ll gao(VI a,ll n,ll mod,bool prime=true) {
VI c;
if(prime) c=BM(a); //素数使用BM
else c=ReedsSloane(a,mod); //合数使用ReedsSloane
c.erase(c.begin());
FOR(i,0,SZ(c)-1) c[i]=(mod-c[i])%mod;
return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}
};
ll qpow(ll a,ll b, ll mod)
{
ll res=1;
while(b)
{
if(b&1) res=res*a%mod;
a=a*a%mod,b>>=1;
}
return res;
}
vector<ll>tmp;
int n, m;
int main()
{
scanf("%d%d", &n,&m);
ll a = 0, b = 1, c, sum = 1;
tmp.push_back(0);tmp.push_back(1);
for(int i = 2;i <= 2*57;i++)
{
c = (a + b) % mod;
a = b;
b = c;
sum = (sum + qpow(c, m, mod)) % mod;
tmp.push_back(sum);
//printf("%lld\n", c);
}
//for(int i = 0;i <= 2*m;i++) printf("%lld\n", tmp[i]);
printf("%lld\n", linear_seq::gao(tmp, n, mod, false));
}
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