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 a[3];
int main(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    for (int i = 0; i < 3; ++i) cin >> a[i];
    bool flag = true;
    if (a[0] != a[1] && a[0] != a[2] && a[1] != a[2]) flag = false;
    int mx = a[0];
    int mi = a[0];
    for (int i = 0; i < 3; ++i) {
      mx = max(mx, a[i]);
      mi = min(mi, a[i]);
    }
    int cnt = 0;
    for (int i = 0; i < 3; ++i) {
      if (a[i] == mx) ++cnt;
    }
    if (cnt < 2) flag = false;
    if (!flag)
      cout << "NO" << '\n';
    else {
      cout << "YES" << '\n';
      cout << mx << ' ' << mi << ' ' << mi << '\n';
    }
  }
  return 0;
}