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;
long long MOD = 1e9 + 7;
const int INF = 2e9 + 10;
void solve() {
  vector<int> a(3);
  int mx = 0, notmx = 0, c = 0;
  for (int &x : a) {
    cin >> x;
    mx = max(x, mx);
  }
  for (int x : a) {
    if (x == mx) {
      c++;
    } else {
      notmx = x;
    }
  }
  if (c < 2) {
    cout << "NO" << endl;
    return;
  } else if (c == 3) {
    cout << "YES" << endl;
    cout << mx << " " << mx << " " << mx << endl;
  } else {
    cout << "YES" << endl;
    cout << mx << " " << notmx << " " << notmx << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}