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:
for _ in range(int(input())):
    x,y,z=map(int,input().split())
    if (x==y and z<y):
        print('YES')
        print(x,z,z,sep=' ')
    elif (y==z and x<y):
        print('YES')
        print(y,x,x,sep=' ')
    elif (x==z and y<x):
        print("YES")
        print(x,y,y,sep=' ')
    elif (x==y==z):
        print("YES")
        print(x,y,z,sep=' ')
    else:
        print('NO')

'''
#include <bits/stdc++.h>
using namespace std;
#define ll long long

int main() {
	int t;
	cin>>t;
	while(t--){
int x,y,z;
cin>>x>>y>>z;
if(x==y && y==z){
	cout<<"YES"<<endl;
	cout<<x<<" "<<y<<" "<<z<<endl;
}
else if(x==y && z<x){
		cout<<"YES"<<endl;
	cout<<x<<" "<<z<<" "<<z<<endl;
}
else if(x==z && y<x){
		cout<<"YES"<<endl;
	cout<<x<<" "<<y<<" "<<y<<endl;
}
else if(y==z && y>x){
		cout<<"YES"<<endl;
	cout<<y<<" "<<x<<" "<<x<<endl;
}
else{
	cout<<"NO"<<endl;
}

   }
	return 0;
}
'''