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:
# cook your dish here
#code
import math
import collections
from sys import stdin,stdout,setrecursionlimit
from bisect import bisect_left as bsl
from bisect import bisect_right as bsr
import heapq as hq
setrecursionlimit(2**20)

t = 1
t = int(stdin.readline())

for _ in range(t):
    #n = int(stdin.readline())
    #s = str(stdin.readline().strip('\n'))
    x,y,z = list(map(int, stdin.readline().rstrip().split()))
    d = {}
    #d[x]=1;d[y]=1;d[z]=1;
    
    for i in [x,y,z]:
        if(i not in d):
            d[i]=0
        d[i]+=1
    if(len(d)==1):
        print('YES')
        print(str(x)+' '+str(x)+' '+str(x))
    elif(len(d)==3):
        print('NO')
    else:
        mi = min(x,y,z)
        mx = max(x,y,z)
        if(d[mi]==2):
            print('NO')
        else:
            if(True):
                print('YES')
                print(str(mi)+' '+str(mi)+' '+str(mx))
            else:
                print('NO')