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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int q = 0; q < t; q++) {
    int x, y, z;
    cin >> x >> y >> z;
    if (x == y && y == z && z == x) {
      cout << "YES\n";
      cout << x << " " << y << " " << z << "\n";
    } else if (x != y && y != z && z != x) {
      cout << "NO\n";
    } else if (x == y && y != z) {
      int d = x;
      if (z > d) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        cout << d << " " << z << " " << z << "\n";
      }
    } else if (x != y && y == z) {
      int d = y;
      if (x > d) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        cout << x << " " << x << " " << d << "\n";
      }
    } else if (x == z && z != y) {
      int d = x;
      if (y > d) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        cout << y << " " << d << " " << y << "\n";
      }
    }
  }
  return 0;
}