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 int mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long int test = 1;
  cin >> test;
  while (test--) {
    long long int n = 3;
    long long int a[n];
    cin >> a[0] >> a[1] >> a[2];
    sort(a, a + n);
    long long int A = 1;
    long long int B = a[0];
    long long int C = a[2];
    long long int b[n];
    b[0] = max(A, B);
    b[1] = max(A, C);
    b[2] = max(B, C);
    sort(b, b + n);
    int f = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        f = 1;
        break;
      }
    }
    if (f)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      cout << A << ' ' << B << ' ' << C << endl;
    }
  }
  return 0;
}