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).

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

2 16
3 27
7 4357186184021382204544

4
3
1234

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 1010
int n;

struct LL {
	int len, A[maxn];
	LL() { len = 1; memset(A, 0, sizeof(A)); }
	LL operator = (const int& t) {
		len = 1; A[1] = t;
		while(A[len] > 9) A[len+1] += A[len] / 10, A[len] %= 10, len++;
		return *this;
	}
	LL operator = (const LL& t) {
		len = t.len;
		memset(A, 0, sizeof(A));
		for(int i = 1; i <= len; i++) A[i] = t.A[i];
		return *this;
	}
	LL operator + (const LL& t) const {
		LL ans; ans.len = max(len, t.len);
		for(int i = 1; i <= len; i++) ans.A[i] = A[i];
		for(int i = 1; i <= t.len; i++) ans.A[i] += t.A[i];
		for(int i = 1; i < ans.len; i++) ans.A[i+1] += ans.A[i] / 10, ans.A[i] %= 10;
		while(ans.A[ans.len] > 9) ans.A[ans.len+1] += ans.A[ans.len] / 10, ans.A[ans.len] %= 10, ans.len++;
		return ans;
	}
	LL operator * (const LL& t) const {
		LL ans; ans.len = len + t.len - 1;
		if((len == 1 && !A[1]) || (t.len == 1 && !t.A[1])) return ans = 0;
		for(int i = 1; i <= len; i++)
			for(int j = 1; j <= t.len; j++)
				ans.A[i+j-1] += A[i] * t.A[j];
		for(int i = 1; i < ans.len; i++) ans.A[i+1] += ans.A[i] / 10, ans.A[i] %= 10;
		while(ans.A[ans.len] > 9) ans.A[ans.len+1] += ans.A[ans.len] / 10, ans.A[ans.len] %= 10, ans.len++;
		return ans;
	}
	LL operator *= (const LL& t) {
		*this = *this * t;
		return *this;
	}
	LL operator / (const int& t) const {
		LL ans; int f = 0;
		for(int i = len; i; i--) {
			f = f * 10 + A[i];
			if(f >= t) ans.len = max(ans.len, i);
			ans.A[i] = f / t; f %= t;
		}
		return ans;
	}
	bool operator < (const LL& t) const {
		if(len != t.len) return len < t.len;
		for(int i = len; i; i--) if(A[i] != t.A[i]) return A[i] < t.A[i];
		return 0;
	}
	void print() {
		for(int i = len; i; i--) putchar(A[i] + ‘0‘); putchar(‘\n‘);
		return ;
	}
} p;

LL Lread() {
	LL x, f, ten; x = 0; f = 1; ten = 10; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)) {
		LL tmp; tmp = (int)(c - ‘0‘);
		x = x * ten + tmp; c = getchar();
	}
	return x * f;
}

LL Pow(LL x, int a) {
	LL ans; ans = 1;
	for(int i = 1; i <= a; i++) {
		ans *= x;
		if(p < ans) return ans;
	}
	return ans;
}

int main() {
	while(scanf("%d", &n) == 1) {
		p = Lread();
		LL l, r, one, l1; one = 1; l = 0; l1 = 0; r = p + one;
		while(l + one < r) {
			LL mid; mid = (l + r) / 2;
			LL tmp = Pow(mid, n);
			if(p < tmp) r = mid; else l = mid;
		}
		
		l.print();
	}
	
	return 0;
}