Codeforces Round #326 (Div. 2) B

1.每一个合数都可以由若干个素数相乘而得到

2.质因数知识 ：求一个数因数的个数等于它的每个质因数的次数加一的和相乘的积
因为质因数可以不用,所以要加一.
例如6=2x3,两个质因数都是一次,如果两个质因数都不用,它的一个因数是1；只用因数2,它的第二个因数就是2；只用因数3,它的第三个因数就是3；两个质因数都用,它的第四个因数就是6。

所以6的因数的个数就是(1+1)(1+1)=4。
再如12=2²x3,两个质因数,2是2次,3是一次,如果两个质因数都不用,它的一个因数是1；只用一个因数2,它的第二个因数就是2；两个2都用,它的第三个因数就是4；不用3时就有3个因数,再用上3时就是(2+1)(1+1)=6个因数了。

然后下面这道题的答案就是每个质因数都只乘一次     10就是2x5。12就是2x3。

B. Duff in Love

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Duff is in love with lovely numbers! A positive integer x is called lovely if and only if there is no such positive integer a > 1 such that a2 is a divisor of x.

Malek has a number store! In his store, he has only divisors of positive integer n (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.

Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.

Input

The first and only line of input contains one integer, n (1 ≤ n ≤ 1012).

Output

Print the answer in one line.

Examples

input

10

output

10

input

12

output

6

Note

In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.

In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 22, so 12 is not lovely, while 6 is indeed lovely.

AC代码

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>
using namespace std;
const int maxn= 510;
const int maxm= 1e4+10;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int main()
{
    ll n;
    while(scanf("%I64d",&n)!=EOF)   //codeforces 不支持%lld
    {
        ll sum=1;
        for(int i=2;i<=sqrt(n);i++)     //要用sqrt函数 用i*i<=n 要把i定义为ll i*i会爆int
        {
            if(n%i==0)
                sum*=i;
            while(n%i==0)          //除去倍数
                n/=i;
        }
        printf("%I64d
",sum*n);
    }
    return 0;
}