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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    #include <limits.h>
     
     
    int main(void){
    	
    	long long n0, a, res = 1;
    	static long long arr[100000000][2];
		
    	
    	scanf("%lld", &a);
    	
    	scanf("%lld", &n0);
		
		if(n0 == 1){
			printf("1\n");
		}
		
		long long limit = ((long long)pow(n0, 1.0/a)) * 2;
		
        long long   n;   // 整数
        long long   ns=2;   // 素因数
        long long   j;      // 乗数
    	long long i = 0;
    	
    	arr[i][1] = 1;
    	arr[i][2] = LLONG_MAX;
    	
    	//printf("%lld\n", n0%3);
    	
        for( n=n0,ns=2; limit>=ns; ns++ )
          {
            for( j=0; n%ns==0; j++ )            // 割り切れる間繰り返す
             {
               n/=ns;
             }
    		 
    		 //printf("%d:%d\n", ns, j);
    		 
            if ( j==0 ) continue;               // 1回も割り切れなかった
     
            arr[i][1] = ns;
    		arr[i][2] = j;
    		i++;
         }
    	 
    	 i--;
    	 
    	 for(; i >= 0; i--){
    	 	if(a <= arr[i][2]){
    			res *= arr[i][1];
    	 	}
    	 }
    	 
    	 printf("%lld\n", res);
    	 
    	return 0;
    }