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;
const double eps = 1e-9;
const double PI = acos(-1.0);
const long long int mod = 1e9 + 7;
const int MAXN = 1e6 + 5;
void cp() {
  int x, y, z;
  cin >> x >> y >> z;
  if (x == y || y == z || x == z) {
    int a[] = {x, y, z};
    sort(a, a + 3);
    a[1] = 1;
    sort(a, a + 3);
    do {
      if (max(a[0], a[1]) == x && max(a[0], a[2]) == y &&
          max(a[1], a[2]) == z) {
        cout << "YES\n";
        cout << a[0] << " " << a[1] << " " << a[2] << '\n';
        return;
      }
    } while (next_permutation(a, a + 3));
    cout << "NO\n";
  } else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    cp();
  }
  return 0;
}