求a^b的约数对mod取模

const int maxn=+;

int prime[maxn];
void marktable(int n){
    memset(prime,,sizeof(prime));
    for(int i=;i<=n;i++){
        if(!prime[i]) prime[++prime[]]=i;
        for(int j=;j<=prime[]&&prime[j]<=n/i;j++){
            prime[prime[j]*i]=;
            if(i%prime[j]==) break;
        }
    }
}
long long factor[][];
int fatCnt;
int getFactors(long long x){
    fatCnt=;
    long long tmp=x;
    for(int i=;prime[i]<=tmp/prime[i];i++){
        factor[fatCnt][]=;
        if(tmp%prime[i]==){
            factor[fatCnt][]=prime[i];
            while(tmp%prime[i]==){
                factor[fatCnt][]++;
                tmp/=prime[i];
            }
            fatCnt++;
        }
    }
    if(tmp!=) {
        factor[fatCnt][]=tmp;
        factor[fatCnt++][]=;
    }
    return fatCnt;
}
int mod,A,B;
template<class T,class T1> T fast_mod(T a,T b,T1 Mod){
    a%=mod;
    if(b==) return ;
    T ans=,base=a;
    while(b!=){
        if(b&)ans=(ans*base)%Mod;
        base=(base*base)%Mod;
        b>>=;
    }
    return ans;
}

long long sum(long long p,long long n){
    if(p==) return ;
    if(n==) return ;
    if(n&) return ((+fast_mod(p,n/+,mod))%mod*sum(p,n/)%mod)%mod;
    else return ((+fast_mod(p,n/+,mod))%mod*sum(p,n/-)+fast_mod(p,n/,mod)%mod)%mod;
}
long long solve(long long A,long long B){
    getFactors(A);
    long long ans=;
    for(int i=;i<fatCnt;i++){
        ans*=sum(factor[i][],B*factor[i][])%mod;
        ans%=mod;
    }
    return ans;
}