A permutation p of size n is an array such that every integer from 1 to n occurs exactly once in this array.

Let's call a permutation an almost identity permutation iff there exist at least n - k indices i (1 ≤ i ≤ n) such that pi = i.

Your task is to count the number of almost identity permutations for given numbers n and k.

Input
The first line contains two integers n and k (4 ≤ n ≤ 1000, 1 ≤ k ≤ 4).

Output
Print the number of almost identity permutations for given n and k.

Examples
Input
4 1
Output
1
Input
4 2
Output
7
Input
5 3
Output
31
Input
5 4
Output
76

题意：
给你一个整数n和k，问有多少个n的全排列中 下标和数值相等的个数是大于等于n-k 的。
思路：
那么我们不妨从n-k到n枚举 下标和数值相等的个数i，对于每一个i，我们可以从n中选择i个数，让他们在1，2,3,4，，，n这个排列中位置不变，剩下的n-i个数，不在自己的位置上，那么问题就转化为 对于每一个i，求C(n,i)a(i)，a(i)是i个数的全排列，每一个数都不在原来位置上的排列种类数。 而a[i] 是一个 递推数列 a[i]=ia[i-1 ]+-1^i ，我们知道a[0]=1，这样顺推就可以求出所有i对答案的贡献，累加起来就是答案值了。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll a[maxn];

ll C(ll m,ll n)//m 中 选 n 个
{
    long long ans=1;
    for(long long k=1; k<=n; k++)
    {
        ans=(ans*(m-n+k))/k;
    }
    return ans;
}

int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    a[0]=1ll;
    a[1]=0ll;
    a[2]=1ll;
    a[3]=2ll;
    a[4]=9ll;
    int base=-1;
    repd(i,5,1010)
    {
        a[i]=1ll*i*a[i-1]+base;
        base*=-1;
    }
    ll n,k;
    cin>>n>>k;
    ll ans=0ll;
    repd(i,n-k,n)
    {
        ans+=C(n,i)*a[n-i];
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}