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;
void solve() {
  long long a, b, c;
  cin >> a >> b >> c;
  if ((a == b) && (b == c)) {
    cout << "YES" << endl;
    cout << a << " " << b << " " << c << endl;
  } else {
    unordered_map<long long, long long> mp;
    mp[a]++;
    mp[b]++;
    mp[c]++;
    int flag = 0, x1, x2, x3;
    for (auto i : mp) {
      if (i.second == 2) {
        flag = 1;
        x1 = i.first;
      }
      if (i.second == 1) {
        x2 = i.first;
      }
    }
    if (x1 < x2) {
      cout << "NO" << endl;
    } else if (flag == 0) {
      cout << "NO" << endl;
    } else if (flag == 1) {
      cout << "YES" << endl;
      if (x2 == 1) {
        cout << x1 << " " << x2 << " " << 1 << endl;
      } else {
        cout << x1 << " " << x2 << " " << (x2 - 1) << endl;
      }
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}