Problem: 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 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 %I64d specifier.

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

3
4 5 6


Output

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".

Solution (in PYTHON3):

Solution:
# http://codeforces.com/problemset/problem/230/B
input()
primes = [0] * int(1e6)
s = set()
for i in range(2, int(1e6)):
    if not primes[i]:
        s.add(i*i)
        primes[i*i::i] = [1]*len(primes[i*i::i])

for i in map(int, input().split()):
    print("YES" if i in s else "NO")

'''
Solution 2:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long int x,y,n,i,j;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        scanf("%lld",&x);
        y=sqrt(x);
        for(j=2;j*j<y;j++)
            if(x%j==0) break;
        if(j*j>y&&y*y==x&&x>1) printf("YES\n");
        else printf("NO\n");
    }
}
'''