HDU-4983 Goffi and GCD

$Description$
$\sum_{a=1}^{n}\sum_{b=1}^{n}[gcd(a,n) \cdot gcd(b,n)=n]$

$Solution$
$\sum_{a=1}^{n}\sum_{b=1}^{n}[gcd(a,n) \cdot gcd(b,n)=n] \\=\sum_{d|n}\sum_{a=1}^{n}[gcd(a,n)=d]\sum_{b=1}^{n}[gcd(b,n)=\frac{n}{d}] \\=\sum_{d|n}\sum_{a=1}^{\frac{n}{d}}[gcd(a,\frac{n}{d})=1]\sum_{b=1}^{d}[gcd(b,d)=1] \\ =\sum_{d|n}\varphi (\frac{n}{d}) \cdot \varphi(d) \\ =\prod_{p^s||n}\sum_{k=0}^{s}\varphi(p^k) \cdot \varphi(p^{s-k}) \\ =\prod_{p^s||n}\sum_{k=0}^{s}(p^k-p^{k-1}) \cdot (p^{s-k}-p^{s-k-1}) \\ =\prod_{p^s||n}(p^{s-2} \cdot (p-1)^2 \cdot (s-1))[s\geq 2]+2\cdot p^s-2 \cdot p^{s-1}$

$Code$

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=1e6;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
ll mul(ll a,ll b,ll c){ll ans=0;for(;b;b>>=1,a=(a+a)%c)if(b&1)ans=(ans+a)%c;return ans;}
ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=mul(a,a,c))if(b&1)ans=mul(ans,a,c);return ans;}
int main(){
    itn n,k;
    while(~sc("%d%d",&n,&k)){
        if(n==1||k==2){
            printf("1\n");continue;
        }
        if(k>2){
            printf("0\n");continue;
        }
        ll ans=1;
        for(int i=2;1ll*i*i<=n;i++){
            if(n%i==0){
                itn cnt=0;
                ll p=1;
                while(n%i==0)n/=i,cnt++,p*=i;
                if(cnt>=2){
                    ll res=1LL*(i-1)*(i-1)*(cnt-1)%mod+2LL*(i-1)*i%mod;
                    res%=mod;
                    ans=ans*res%mod*(p/i/i)%mod;
                }else {
                    ans=ans*(2LL*(i-1)%mod)%mod;
                }
            }
        }
        if(n>1)ans=ans*2LL*(n-1)%mod;
        printf("%lld\n",ans);
    }
}