#include <fstream>
#define MOD 9901
#define limit 1000005

using namespace std;

ifstream f("sumdiv.in");
ofstream g("sumdiv.out");

long long int a, b, n, k = -1, DA[limit];
bool prim[limit];

void ciur()
{
	for (long long int i = 2; i < limit; ++i) {
		if (prim[i] == 0) {
			DA[++k] = i;

			for (long long int j = i + i; j < limit; j += i) {
				prim[j] = 1;
			}
		}
	}
}

long long int power(long long int x, long long int p)
{
	long long int rez = 1;
	while (p) {
		if (p & 1) rez = (rez*x) % MOD;
		x = (x*x) % MOD;
		p >>= 1;
	}
	return rez % MOD;
}

void solve()
{
	long long int sum = 1, pw, i, pw1, pw2;
	for (i = 0; i <= k && DA[i] * DA[i] <= n; i++) {
		if (n % DA[i]) continue;
		pw = 0;
		while (n % DA[i] == 0) {
			n /= DA[i];
			pw++;
		}
		pw1 = (power(DA[i], pw + 1) - 1) % MOD;
		pw2 = (power(DA[i] - 1, MOD - 2)) % MOD;
		sum = (((sum*pw1) % MOD) * pw2) % MOD;
	}

	if (n > 1) sum = (sum * (n + 1)) % MOD;

	g << sum;
}

int main()
{
	ciur();
	f >> a >> b;
	n = power(a, b) % MOD;
	solve();
	f.close();
	g.close();
	return 0;
}