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>
const int BIG = 1e5 + 55;
const long long MOD = 1e9 + 7;
using namespace std;
int t, a[BIG];
void solve() {
  int x, y, z;
  cin >> x >> y >> z;
  a[0] = x;
  a[1] = y;
  a[2] = z;
  a[3] = 0;
  int A, B, C;
  for (int i = 0; i < 4; i++) {
    A = a[i];
    for (int j = 0; j < 4; j++) {
      B = a[j];
      for (int k = 0; k < 4; k++) {
        C = a[k];
        if (x == max(A, B) && y == max(A, C) && z == max(B, C)) {
          cout << "YES\n" << A << " " << B << " " << C << "\n";
          return;
        }
      }
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}