We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More generally, x is a perfect pth power if, for some integer b, x = bp. Given an integer x you are to determine the largest p such that x is a perfect pth power.

Input

Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing 0 follows the last test case.

Output

For each test case, output a line giving the largest integer p such that x is a perfect p th power.

 

Sample Input 1 Sample Output 1 
17
1073741824
25
0

1
30
2

题目的大概意思是——给出一个n，n=b^p，求出最大p值。（b未知）

思路：

首先利用唯一分解定理，把n写成若干个素数相乘的形式。接下来对于每个素数的指数求最大公约数，该公约数就是所能到达的最大p值。

有一点要注意的是如果n为负数的话，如果当前p值为偶数，就一直除2直到p为奇数。（被坑了。）

 //Asimple
#include <bits/stdc++.h>
#define INF (1<<20)
#define mod 10007
#define swap(a,b,t) t = a, a = b, b = t
#define CLS(a, v) memset(a, v, sizeof(a))
#define debug(a)  cout << #a << " = "  << a <<endl
using namespace std;
typedef long long ll;
const int maxn = 70000;
const double PI=atan(1.0)*4;int n, m, num, res, ans, len, T, k;int a[maxn];
int pr[maxn];

int gcd(int a, int b) {
    return b==0?a:gcd(b, a%b);
}

void solve() {
    CLS(a, 0);
    len = 0;
    for(int i=2; i*i<=maxn; i++) {
        if( !a[i] ) {
            pr[len++] = i;
            for(int j=i; j<maxn; j+=i)
                a[j] = 1;
        }
    }
}

void input() {
    solve();
    while( cin >> n && n ) {
        ans = 0;
        int f = 0;
        if( n < 0 ) {
            n = -n;
            f = 1;
        }
        for(int i=0; i<len && n>1; i++) {
            if( n%pr[i]==0 ) {
                res = 0;
                while( n%pr[i]==0 ) {
                    res ++;
                    n /=pr[i];
                }
                ans = gcd(ans, res);
            }
        }
        if( n>1 ) ans = 1;
        if( f ) while( ans%2==0 ) ans/=2;
        cout << ans << endl;
    } 
}

int main(){
    input();
    return 0;
}