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;
int main() {
  int t;
  cin >> t;
  bool* res1 = new bool[t];
  int* res2 = new int[3 * t];
  for (int f = 0; f < t; f++) {
    int x, y, z;
    cin >> x >> y >> z;
    if (y > x) {
      x += y;
      y = x - y;
      x -= y;
    }
    if (z > x) {
      x += z;
      z = x - z;
      x -= z;
    }
    if (y < z) {
      y += z;
      z = y - z;
      y -= z;
    }
    if (x == y || x == z) {
      res2[f * 3] = x;
      res2[f * 3 + 1] = z;
      if (y == z)
        res2[f * 3 + 2] = z;
      else
        res2[f * 3 + 2] = 1;
      res1[f] = true;
    } else
      res1[f] = false;
  }
  for (int i = 0; i < t; i++)
    if (res1[i])
      cout << "YES\n"
           << res2[i * 3] << " " << res2[i * 3 + 1] << " " << res2[i * 3 + 2]
           << endl;
    else
      cout << "NO\n";
}