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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<long long int> nos;
vector<long long int> tree;
int main() {
  long long int t = 1;
  cin >> t;
  while (t > 0) {
    t--;
    long long int x, y, z;
    cin >> x >> y >> z;
    long long int a = 0;
    long long int b = 0;
    long long int c = 0;
    if (x == y && y == z) {
      cout << "YES" << endl;
      cout << x << " " << y << " " << z << endl;
      continue;
    }
    if (x != y && y != z && x != z) {
      cout << "NO" << endl;
      continue;
    }
    if (x == y && z < x) {
      cout << "YES" << endl;
      cout << x << " " << z << " " << z << endl;
      continue;
    }
    if (z == y && x < y) {
      cout << "YES" << endl;
      cout << x << " " << x << " " << y << endl;
      continue;
    }
    if (z == x && y < z) {
      cout << "YES" << endl;
      cout << x << " " << y << " " << y << endl;
      continue;
    }
    cout << "NO" << endl;
  }
}