CodeForces 230B T-primes

题目衔接：http://codeforces.com/problemset/problem/230/B

B. T-primes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer t Т-prime, if t has exactly three distinct positive divisors.

You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.

Input

The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains nspace-separated integers xi (1 ≤ xi ≤ 1012).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64dspecifier.

Output

Print n lines: the i-th line should contain "YES" (without the quotes), if number xi is Т-prime, and "NO" (without the quotes), if it isn't.

Examples

input

Copy

3
4 5 6

output

Copy

YES
NO
NO

Note

The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

题目大意：现在重新定义一个T素数，这个数只能被1自己和另外一个因子整除，即只有三个因子，现在让你判断是不是T素数

思路：既然只有三个因子，那么第三个一定是它的平方根，而且它的平方根还不能有其他因子，即平方根为素数！，知道了后就直接打表，判断是不是素数即可。

代码： 

/*

*/
#include<map>
#include<set>
#include <vector>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define inf 0x3f3f3f
#define esp 1e-8
#define bug {printf("mmp\n");}
#define mm(a,b) memset(a,b,sizeof(a))
#define T() int test,q=1;scanf("%d",&test); while(test--)
const int maxn=1e6+10;
const double pi=acos(-1.0);
const int N=201;
const int mod=1e9+7;
bool a[maxn];
void ff()
{
    a[1]=true;
    for(int i=2;i<=maxn;i++)
    {
        if(a[i]==false)
        {
            for(int j=i*2;j<=maxn;j+=i)
                a[j]=true;
        }
    }
}

int main()
{
    ff();
    T()
    {
        ll n;
        cin>>N;
        double ans=sqrt(n);
        int m=ans;
        if(m==ans)
        {
            if(!a[m])
                cout<<"YES\n";
            else
                cout<<"NO"<<endl;
        }
        else
            cout<<"NO"<<endl;
    }

    return 0;
}