Problem:
Score : 300 points

Problem Statement

There are N integers a_1, a_2, ..., a_N not less than 1.
The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P.

Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.

Constraints

1 \leq N \leq 10^{12}

1 \leq P \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N P

Output

Print the answer.

Sample Input 1

3 24

Sample Output 1

2

The greatest common divisor would be 2 when, for example, a_1=2, a_2=6 and a_3=2.

Sample Input 2

5 1

Sample Output 2

1

As a_i are positive integers, the only possible case is a_1 = a_2 = a_3 = a_4 = a_5 = 1.

Sample Input 3

1 111

Sample Output 3

111

Sample Input 4

4 972439611840

Sample Output 4

206

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

long long gcd(long long x, long long y);

int main(){
	int n;
	long long p;
	long long num[2];
	long long g;
	long long max = 0;
	scanf("%d %lld",&n,&p);
	long long half = 0;
	long long i = 1;
	num[0] = 1;

	if(n == 1){
		printf("%d\n",p);
	}
	else if(p == 1){
		printf("1\n");
	}
	else{
		half = p/n;
		while(i < half){
			if(p % i == 0){
				num[1] = num[0];
				num[0] = i;
				if(max < gcd(num[0],num[1])){
					max = gcd(num[0],num[1]);
				}
			}
			i++;
		}
	}
	printf("%lld\n",max);
	return 0;
}

long long gcd(long long x, long long y)
{
	long long r, t;

	if(x == 0 || y == 0)
		{
        	fprintf(stderr, "引数エラー\n");
        	return 0;
	}
	if(x<y){
		t = x;
		x = y;
		y = t;
	}

	// ユーグリッドの互除法
	while((r = x % y) != 0)
    	{
        	x = y;
       		y = r;
	}
	return y;
}