97. 约数之和

/*
	A = p1^k1 * p2^k2 * ... * pn^kn
	
	所有的约数的个数:
	(k1 + 1) * (k2 + 1) * ... * (kn + 1)
	
	所有的约数之和:
	(p1^0 + p1^1 + ... + p1^k1) * (p2^0 + p2^1 + ... + p2^k2) * ... *
	(pn^0 + pn^1 + ... + pn^kn)
	
*/

/*
	sum(p, k) = p^0 + p^1 + ... + 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))
			  = (1 + p^(k/2 + 1)) * sum(p, k / 2)
*/

#include <iostream>

using namespace std;

const int mod = 9901;

// 快速幂 
int qmi(int a, int k)
{
	a %= mod;　　// 注意此处要对a模9901
	int res = 1;
	while(k)
	{
		if(k & 1)	res = res * a % mod;
		a = a * a % mod;
		k >>= 1;
	}
	return res;
}

// 等比数列求和 
int sum(int p, int k)
{
	if(k == 0)	return 1;
	if(k % 2 == 0)	return (p % mod * sum(p, k - 1) + 1) % mod;
	return (1 + qmi(p, k/2 + 1)) * sum(p, k / 2) % mod;
}

int main()
{
	int A, B;
	cin >> A >> B;
	int res = 1;
	for(int i = 2; i <= A; ++ i)
	{
		int s = 0;
		while(A % i == 0)
		{
			s ++;
			A /= i;
		}
		if(s)	res = res * sum(i, B * s) % mod;
	}
	if(!A)	res = 0;
	cout << res << endl;
	
	return 0;
}