Problem: You are given three positive (i.e. strictly greater than zero) integers x, y and z.

Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.

You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).

Output

For each test case, print the answer:

  * "NO" in the only line of the output if a solution doesn't exist; 
  * or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order. 

Example

Input


5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000


Output


YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000

Solution (in PYTHON3):

Solution:
from sys import stdin, stdout
cin = stdin.readline
cout = stdout.write
mp = lambda: list(map(int, cin().split()))

t, = mp()
for _ in range(t):
	x, y, z = mp()
	
	if z == y and y >= x:
		cout('YES\n' + str(x) + ' ' + str(x) + ' ' + str(y) + '\n')
	elif x == z and z >= y:
		cout('YES\n' + str(x) + ' ' + str(y) + ' ' + str(y) + '\n')
	elif x == y and x >= z:
		cout('YES\n' + str(x) + ' ' + str(z) + ' ' + str(z) + '\n')
	else:
		cout('NO\n')