/*

     题意：k ^ n = p，求k

     1. double + pow：因为double装得下p，k = pow (p, 1 / n);

     基础知识: 类型    长度 (bit)    有效数字    绝对值范围

             float    32             6~7         10^(-37) ~ 10^38

             double    64            15~16        10^(-307) ~ 10^308

        long double    128              18~19         10^(-4931) ~ 10 ^ 4932

     2. 二分查找：和1类似

     3. 取对数：n*ln(k)=ln(p)    ln(k)=ln(p)/n    k=exp(ln(p)/n)

 */

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <map>

 #include <queue>

 #include <vector>

 using namespace std;

 const int MAXN = 1e6 + ;

 const int INF = 0x3f3f3f3f;

 int main(void)        //POJ 2109 Power of Cryptography

 {

     //freopen ("D.in", "r", stdin);

     double n, p, k;

     while (~scanf ("%lf%lf", &n, &p))

     {

         printf ("%.0f\n", pow (p,  / n));

     }

     return ;

 }

 /*

 #include <cstdio>

 #include <cmath>

 #include <algorithm>

 #include <iostream>

 using namespace std;

 void BinarySearch(int l, int r, double n, double p)

 {

     int mid;

     while (l <= r)

     {

         mid = l + (r - l) / 2;

         double tmp = pow (mid, n);

         if (tmp == p)

         {

             printf ("%d\n", mid);    return ;

         }

         else if (tmp < p)    l = mid + 1;

         else    r = mid - 1;

     }

 }

 int main(void)

 {

     //freopen ("D.in", "r", stdin);

     double n, p;

     while (~scanf ("%lf%lf", &n, &p))

     {

         BinarySearch (1, 1e9, n, p);

     }

     return 0;

 }

 */

 /*

 #include <cstdio>

 #include <cmath>

 #include <algorithm>

 #include <iostream>

 using namespace std;

 int main(void)        //POJ 2109 Power of Cryptography

 {

     //freopen ("D.in", "r", stdin);

     double n, p;

     while (~scanf ("%lf%lf", &n, &p))

     {

         printf ("%.0f\n", exp (log (p) / n));

     }

     return 0;

 }

 */