$\text{令} sum(p,c) = \sum_{i=0}^{c} (p^i)$

$即1+p^1+p^2+...+p^c$

$\text{当}c为偶数时：$

$sum(p,c) = 1 +...+p^{\frac{c}{2}}+p^{\frac{c}{2}+1}+...+p^c$

$\hspace{1.54cm}=1+...+p^{\frac{c}{2}}+p^{\frac{c}{2}+1}(1 + ...+p^{\frac{c}{2} - 1})$

$\hspace{1.54cm}=(1+...+p^{\frac{c}{2}})+p^{\frac{c}{2}+1}(1 + ...+p^{\frac{c}{2} - 1}+p^{\frac{c}{2}}) -p^{c+1}$

$\hspace{1.54cm}=(1 + p^{\frac{c}{2} + 1})\cdot sum(p,\frac{c}{2})-p^{c+1}$

#include <bits/stdc++.h>
using namespace std;
int b;
const int mod = 9901;
struct node{
	int p = 0,c = 0;
	node (int _p,int _c) : p(_p),c(_c){};
};
vector<node> v;
int qmi(int a,int n){
	int res = 1;
	while(n){
		if(n & 1) res = res * a % mod;
		n >>= 1;
		a = a * a % mod;
	}
	return res;
}
void cal(int a){
	for(int i=2;i<=sqrt(a);i++){
		if(a % i == 0){
//			node n = {a,0};
			v.push_back({i,0}); 
			while(a % i == 0){
		 		a /= i;
				v.back().c ++;
			}
			v.back().c *= b;
		}
	}
	if(a > 1){
		v.push_back({a,b});
	}
}
int sum(int p,int c){
	if(!c) return 1;
	if(c & 1){
		return ((1 + qmi(p,(c + 1) / 2)) % mod) * (sum(p,(c - 1) / 2) % mod);
	}else{
		return ((1 + qmi(p,(c / 2) + 1)) % mod) * (sum(p,c / 2) % mod) - qmi(p,c + 1);
        //return ((1 + qmi(p,c / 2) % mod) * ((sum(p,c / 2 - 1) % mod)) + qmi(p,c));改成这句就对了
	}
}
signed main(){
	int a,ans = 1;
	cin>>a>>b;
	if(a == 0){
		cout<<0;
		return 0;
	}
	cal(a);
/*	for(int i=0;i<v.size();i++){
		cout<<v[i].p<<' '<<v[i].c<<endl;
	}*/
	for(int i=0;i<v.size();i++){
//		cout<<v[i].p<<' '<<v[i].c<<endl;
		ans *= sum(v[i].p,v[i].c) % mod;
		ans %= mod;
	}
	cout<<ans;
	return 0;
}