约数之和

#include <bits/stdc++.h>
using namespace std;


const int mod = 9901;

int qmi(int a, int k)
{
    a %= mod;
    int res = 1;
    while(k)
    {
        if(k & 1)
            res =  res * a % mod;
        a = a * a % mod;
        k >>= 1;
    }
    return res;
}


//p ^ 0 + ... + p ^ k
int sum(int p, int k)
{
    if(k == 0) return 1;
    if(k % 2 == 0)
    {
        //当k是偶数的时候：p * (p ^ 0 + ... + p ^ k - 1) = p ^ 1 + ... + p ^ k
        return (p % mod * sum(p, k - 1) + 1) % mod;
    }
    //当k是奇数的时候
    //p ^ 0 + ... + p ^ k = (p ^ 0 + ... + p ^ (k / 2)) + (p ^ (k / 2 + 1) + ... + p ^ k)
    //                    = (p ^ 0 + ... + p ^ (k / 2)) + p ^ (k / 2 + 1) * (p ^ 0 + ... + p ^ (k / 2)) 奇数下取整
    //                    = (1 + p ^ (k / 2) + 1) * sum(p, k / 2)
    return (1 + qmi(p, k/2 + 1)) * sum(p, k / 2) % mod;
}


int main()
{
    int a, b;
    cin >> a >> b;
    int res = 1;
    
    //试除法寻找质数因子
    for(int i = 2; i <= a; i++)
    {
        int s = 0;
        while(a % i == 0)
        {
            s++;
            a /= i;
        }
        if(s)
        {
            res = res * sum(i, s * b) % mod;
        }
    }
    
    if(!a) res = 0;
    cout << res << endl;
    
    return 0;
}