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:
from math import sqrt, ceil
t = int(input())
arr = []
primes = []
m = -1
a = list(map(int, input().split()))
for i in a:
    if (i > m):
        m = i
sieve = []
#print(a)
def Sieve(m):
    target = 1000001
    for i in range(target+1):
        sieve.append(1)
    sieve[0] = sieve[1] = 0
    tar = 10001
    for i in range(2, tar + 1):
        if (sieve[i] == 1):
            for k in range(i*i, target + 1, i):
                sieve[k] = 0
    
    #for i in range(2, target+1):
    #    if (sieve[i] == 1):
    #        primes.append(i)
        
        
Sieve(m)


l = len(primes)
for i in a:
    if (i == 4):
        print("YES")
        continue
    if (i % 2 == 0):
        print("NO")
        continue
    n = i**0.5
    if (int(n) < n):
        print("NO")
    elif (sieve[int(n)] == 1):
        print("YES")
    else:
        print("NO")