耍假第一周 Q

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 n space-separated integers x_i (1 ≤ x_i ≤ 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 %I64d specifier.

Output

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

Example

Input

34 5 6

Output

YESNONO

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".

//Q************
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn=1000000+5;
int a[maxn+1];
void isprime()
{
    a[0]=a[1]=1;
    memset(a,0,sizeof(a));
    for(int i=2;i<=maxn;i++){
        if(!a[i]){
            for(int j=i+i;j<=maxn;j+= i){
                a[j]=1;
            }
        }
    }
}
int main()
{
    int t;
    cin>>t;
    isprime();
    while(t--){
        ll n;
        cin>>n;
        ll x=sqrt(n)+0.5;
        if(x>1&&x*x==n&&a[x]==0){
           cout<<"YES"<<endl;
        }else {
            cout<<"NO"<<endl;
        }
    }
    return 0;
}