/*#include <stdio.h>
#include <string.h>

const int n_max = 10001; // Definim numarul maxim de cifre al numerelor
const int m = 1999999973;

int main()
{
    unsigned int i, n, p;
    long long a, sol = 1;

    freopen("lgput.in","r",stdin);
    freopen("lgput.out","w",stdout);
    scanf("%d %d", &n, &p);
    a = n;
    for (i = 0; (1<<i) <= p; ++ i)  // Luam toti biti lui p la rand
    {
        if ( ((1<<i) & p) > 0) // Daca bitul i din p este 1 atunci adaugam n^(2^i) la solutie
            sol= (sol * a) % m;

            a=(a * a) % m; // Inmultim a cu a ca sa obtinem n^(2^(i+1))
    }
    printf("%lld\n", sol); // Afisam solutia
}

*/#include <fstream>
//#include <iostream>

using namespace std;

ifstream fin("lgput.in");
ofstream fout("lgput.out");

long long putere(long long, long long);

long long a, b, c;

int main()
{
    fin >> a >> c;
    fout << putere(a, c)%1999999973;
    return 0;
}


long long putere(long long a, long long exp)
{
    if(exp <= 1)
    {
        if(exp == 1)
        return a%1999999973;
        if(exp == 0)
            return 1;
    }
    else
    {
        b = putere(a, exp/2)%1999999973;
        if(exp%2 == 0)
            return (((b%1999999973)*(b%1999999973))%1999999973);
        else
            return ((((b%1999999973)*(b%1999999973))*(a%1999999973))%1999999973);
    }
}