E. Number With The Given Amount Of Divisors

E. Number With The Given Amount Of Divisors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.

Input

The first line of the input contains integer n (1 ≤ n ≤ 1000).

Output

Output the smallest positive integer with exactly n divisors.

Examples

input

4

output

6

input

6

output

12

 思路：反素数；

（1）一个反素数的所有质因子必然是从2开始的连续若干个质数，因为反素数是保证约数个数为的这个数尽量小

（2）同样的道理，如果，那么必有

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<queue>
#include<math.h>
#include<set>
#include<vector>
#include<string.h>
using namespace std;
typedef long long LL;
bool prime[100];
int ans[100];
LL minn = 1e18+10;
LL maxx = 1e18;
void  dfs(LL n,int cn,int pre,int need,int cnt);
int main(void)
{
    LL m,n;
    int i,j;
    int cn = 0;
    memset(prime,0,sizeof(prime));
    for(i = 2 ; i < 100 ; i++)
    {
        if(!prime[i])
        {
            ans[cn++] = i;
            for(j = i; (i*j) <= 100 ; j++)
            {
                prime[i*j]=true;
            }
        }
    }
    int ak = 16;
    while(scanf("%lld",&n)!=EOF)
    {
        minn = 1e18+10;
        dfs(1,0,63,n,1);
        printf("%lld
",minn);
    }
    return 0;
}
void  dfs(LL n,int cn,int pre,int need,int cnt)
{
    if(cnt>need)return ;
    if(cnt == need)
    {
         minn = min(minn,n);
        return ;
    }
    LL ap = 1;
    int i,j;
    for(i = 1; i <= pre; i++)
    {   if(maxx/ap>=ans[cn])
        ap*=ans[cn];
        else return ;
        if(maxx/n >= ap)dfs(n*ap,cn+1,pre,need,cnt*(1+i));
        else return ;

    }
}