//Dandu-se doua numere naturale N si P, se cere sa se calculeze restul impartirii lui N^P la 1999999973.
#include <stdio.h>

unsigned long int N, P, mod = 1999999973;

unsigned long int logPow(unsigned long int n, unsigned long int p){
	long int z = 1;

	n = n % 1999999973;
	while(p > 0){
		if(p % 2 == 1){
			z = (z * n) % 1999999973;
		}

		n = (n * n) % 1999999973;
		p = p / 2;
	}
	
	return z;
}

void read(){
	freopen("lgput.in", "r", stdin);
	scanf("%lu", &N);
	scanf("%lu", &P);
}

int main(){
	freopen("lgput.out", "w", stdout);
	read();
	printf("%lu", logPow(N, P));
	return 0;
}