Problem M

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest.
This problem involves the efficient computation of integer roots of numbers.
Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the n ^th. power, for an integer k (this integer is what your program must find).

Input

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10¹⁰¹ and there exists an integer k, 1<=k<=10⁹ such that kⁿ = p.

Output

For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.

Sample Input

   
   
   
   

    
    
    
    2 16
3 27
7 4357186184021382204544
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    4
3
1234
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    很简单，一开始是担心数据过大或者运算次数太多会崩溃。但多虑了。
   
   
   
   
   
   
   
   

    
    
    
    首先，题目的意思是给出指数，给出幂，求指数。其中幂的范围是1<=p<=10^101，而double类型的取值范围是
    
    
    
    4.94065645841246544E-324 到 1.79769313486231570E+308，故p不会溢出。
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    再利用pow(double n.double m)韩式就可。值得注意的是，这里的n代表幂，而m代表指数的倒数。即n^m = x 而x ^ (1/m) = n.