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:
hat = int(input())
lst = [int(x) for x in input().split()]
limit = 1000000
def Prime():
    primeFlag = [True] * limit
    primeFlag[0] = primeFlag[1] = False
    for i in range(2, limit):
        if primeFlag[i]:
            for j in range(i*i, limit, i):
                primeFlag[j] = False
    return primeFlag


primeflag = Prime()
#print(primeflag1)
for i in lst:
    if i == 4:
        print("YES")
    elif i % 2 == 0 or i == 1:
        print("NO")
    else:
        sq = i**0.5
        if sq == int(sq):
            if primeflag[int(sq)] == True:
                print("YES")
            else:
                print("NO")
        else:
            print("NO")
#print(976197352729**0.5)
"""
for i in lst:
    if ran[int(sqrt(i))] == 1:
        print("YES")
    else:
        print("NO")
1
999966000289
"""