#include<iostream>
#include<unordered_map>
using namespace std;
//求等比数列前n项和
//sum(p,k)=p^0+p^1+....+p^k
//当k为奇数时可以划分为:(p^0+p^1+....+p^k/2)+(p^(k/2+1)+p^(k/2+2)+....+p^k),将p^k/2+1提出来.
//                    =(p^0+p^1+....+p^k/2)+p^k/2(p^1+p^2+...p^k/2)=(1+p^k/2+1)*sum(p,k/2);
//奇数项可以套公式
//当k为偶数时想办法转化为奇数求解.sum(p,k)=(p^0+p^1+...+p^k)=p*(p^0+p^1+....+p^k-1)+1=p*sum(p,k-1)+1;

const int mod = 9901;

unordered_map<int,int>prime;

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

int sum(int p,int k)
{
    if(k==0)return 1;
    if(k%2==0)//偶数
    {
        return (p%mod*sum(p,k-1)+1)%mod;
    }
    else return (1+qmi(p,k/2+1))*sum(p,k/2)%mod;

}
void divide (int n)
{
    for(int i=2;i<=n/i;i++)
    {
        while(n%i==0)
        {
            n/=i;
            prime[i]++;
        }
    }
    if(n>1)prime[n]++;
}
int main()
{   
    int a,b;
    cin>>a>>b;
    int res=1;
    divide(a);
    for(auto x:prime)
    {
        int p=x.first,k=x.second;

        res=res*sum(p,k*b)%mod;
    }
    if(!a)res=0;
    cout<<res<<endl;


    return 0;
}