#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string>
#include <string.h>
#define pb push_back

#define SS(a,b) scanf("%d%d",&a,&b);
#define S(a) scanf("%d",&a);
#define SSL(a,b) scanf("%lld%lld",&a,&b);
#define SL(a) scanf("%lld",&a);
#define SSS(a,b,c) scanf("%d %d %d",&a,&b,&c);
#define GI ({int t;scanf("%d",&t);t;})
#define GL ({ll t;scanf("%lld",&t);t;})
#define MAXN 500000
#define FOR(i,n) for(int i=0;i<n;i++)
#define disvec(v) { for(int vec_index=0;vec_index<v.size();vec_index++) cout<<v[vec_index]<<" "; cout<<endl;}
#define TESTIN freopen("input.txt","r",stdin);
#define TESTOUT freopen("output.txt","w",stdout);

#define WRITETEST freopen("input.txt","w",stdout);

using namespace std;
typedef  long long LL;
typedef  long long ll;
typedef pair<LL,LL> PLL;
LL mod=9901LL;
long long  modular(long long  a,long long b,long long c){
    //Calculate a power b mod c
    long long x=1,y=a;
    while(b>0){
         if(b&1) x=(x*y)%c;
         y=(y*y)%c;
         b>>=1;
    }
    return (x%c);
}
//PLL -- > primefactor,power
// Sum of Divisors of a number when number is represented as prime factors
// N=p1^a * p2^ b *p3^c
// =(p1^(a+1))/(p1-1) *(p2^(b+1))/(p2-1)....
vector<PLL > factorize(LL a){
    vector<PLL >temp;
    for(ll x=2;a>=2;x++){
		   if(a<2)break;
           if(a%x==0){
			  LL c=0;
			  while(a%x==0) c++,a/=x;
			  temp.pb(make_pair(x,c));
		   }
    }
    return temp;
}
int main(){
    LL a,b;
    cin>>a>>b;
    vector<PLL > temp=factorize(a);
    /*for(int i=0;i<temp.size();i++)
    	cout<<temp[i].first<<" "<<temp[i].second<<endl;*/
    LL ans=1;
	for(int i=0;i<temp.size();i++){
		LL num=temp[i].first;
		LL cnt=temp[i].second;
		LL totalpower=cnt*b;
		LL first=modular(num,totalpower+1,mod);
		first--;
		// Division by (num-1)  and mod is prime
		// now Let a=(num-1)
		// then multiply by a^(mod-2)%mod
		LL inverse=modular(num-1,mod-2,mod);
		ans*=(first*inverse)%mod;
		ans%=mod;
		if(ans<0)ans+=mod;
	}
	cout<<ans<<endl;
    GI;
    return 0;
}