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 gcd(int a, int b) { return !a ? b : gcd(b % a, a); }
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    vector<int> v(3);
    cin >> v[0] >> v[1] >> v[2];
    set<int> s{v[0], v[1], v[2]};
    int max_val = max(v[0], max(v[1], v[2]));
    if (count((v).begin(), (v).end(), max_val) < 2) {
      cout << "NO";
    } else {
      cout << "YES" << '\n';
      if ((int)s.size() == 1) {
        int p = 3;
        while (p--) {
          for (auto it : s) {
            cout << it << ' ';
          }
        }
      } else {
        for (auto it : s) cout << it << ' ';
        cout << 1;
      }
    }
    if (t) cout << '\n';
  }
}