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:
import math

prime = []
def isprime2(n):
    prime = [True for i in range(n + 1)]
    lis = []
    p = 2
    while (p * p <= n):

        # If prime[p] is not changed, then it is a prime
        if (prime[p] == True):

            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    prime[0] = False
    prime[1] = False

    return prime
    # Print all prime numbers
    for p in range(n + 1):
        if prime[p]:
            lis.append(p)
    return lis


p = isprime2(1000000)
# print(p[7])

# def binarySearch(l, r, x):
#     while l <= r:
#         mid = l + (r - l) // 2;
#         # Check if x is present at mid
#         if lis[mid] == x:
#             return True
#
#             # If x is greater, ignore left half
#         elif lis[mid] < x:
#             l = mid + 1
#
#         # If x is smaller, ignore right half
#         else:
#             r = mid - 1
#
#     # If we reach here, then the element was not present
#     return False


a = int(input())
x = input().split()
count = 0
# e = len(lis)
for l in x:
    f = math.sqrt(int(l))

    # if binarySearch(0, e - 1, int(f)) and f.is_integer():
    #     print("YES")
    # else:
    #     print("NO")

    if f.is_integer() and p[int(f)] == True:
        print("YES")
    else:
        print("NO")