You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).

Output

For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.

Otherwise, print "YES" and any possible such representation.

Example

Input


5
64
32
97
2
12345


Output


YES
2 4 8 
NO
NO
NO
YES
3 5 823 
import math
def getnumbers(n): 
      divisor=[]
      while n % 2 == 0: 
        divisor.append(2), 
        n = n / 2
          
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used 
      for i in range(3,int(math.sqrt(n))+1,2): 
          
        # while i divides n , print i ad divide n 
        while n % i== 0: 
            divisor.append(int(i)), 
            n = n / i 
              
    # Condition if n is a prime 
    # number greater than 2 
      if n > 2: 
          divisor.append(int(n))
    # Initialize the variables with 1 
      a, b, c, size = 0, 0, 0, 0
      a = b = c = 1
      size = len(divisor) 
  
      for i in range(size): 
  
        # check for first number a 
        if (a == 1): 
            a = a * divisor[i] 
  
        # check for second number b 
        elif (b == 1 or b == a): 
            b = b * divisor[i] 
  
        # check for third number c 
        else: 
            c = c * divisor[i] 
    # check for all unwanted condition 
      if (a == 1 or b == 1 or c == 1
        or a == b or b == c or a == c): 
        print('NO') 
      else: 
        print("YES")
        print(a, b, c) 
  
# Driver function 
for i in range(int(input())): 
    n =int(input())
    getnumbers(n)
    


