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 
t = int(input().strip())
import math  
  
  
def primeFactors(n): 
    ans = []
    # Print the number of two's that divide n 
    while n % 2 == 0: 
        ans.append(2) 
        n = n // 2
          
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used
    result = int(math.sqrt(n))
    for i in range(3,result+1,2): 
          
        # while i divides n , print i ad divide n 
        while n % i== 0: 
            ans.append(i) 
            n = n // i 
              
    # Condition if n is a prime 
    # number greater than 2 
    if n > 2: 
        ans.append(n) 
    return ans

for _ in range(t):
    n = int(input().strip())
    ans = primeFactors(n)
    ts = set(ans)
    if len(ts) == 3 and len(ts) == len(ans):
        a = ans[0]
        b = ans[1]
        c = 1
        for i in range(2,len(ans)):
            c *= ans[i]
        print('YES')
        print(a,b,c)
    elif (len(ts) >= 2 and len(ans) >= 4):
        a = ts.pop()
        b = ts.pop()
        c = 1
        for i in range(len(ans)):
            if ans[i] == a:
                ans[i] = 1
                break
        for i in range(len(ans)):
            if ans[i] == b:
                ans[i] = 1
                break
        for i in range(len(ans)):
            c *= ans[i]
        print('YES')
        print(a,b,c)
    elif (len(ts) >= 1 and len(ans) >= 6):
        a = ans[0]
        b = ans[1] * ans[2]
        c = 1
        for i in range(3,len(ans)):
            c *= ans[i]
        print('YES')
        print(a,b,c)
    else:
        print('NO')
    
