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 int N = 2e5 + 5;
const int MOD = 1e9 + 7;
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int x, y, z;
    cin >> x >> y >> z;
    if (x == y && y == z)
      cout << "YES\n" << x << " " << x << " " << x;
    else if (x != y && y != z && z != x)
      cout << "NO";
    else {
      int mx = max(x, max(y, z));
      int mn = min(x, min(y, z));
      int ct[2] = {0, 0};
      int d[3] = {0, 0, 0};
      if (x == mx) d[0] = 1, ct[0]++;
      if (y == mx) d[1] = 1, ct[0]++;
      if (z == mx) d[2] = 1, ct[0]++;
      if (x == mn) ct[1]++;
      if (y == mn) ct[1]++;
      if (z == mn) ct[1]++;
      if (ct[0] > ct[1]) {
        cout << "YES\n";
        if (d[0] && d[1]) d[0] = 0;
        if (d[2] && d[1]) d[1] = 0;
        if (d[0] && d[2]) d[2] = 0;
        for (int i = 0; i < 3; i++) {
          if (d[i])
            cout << mx << " ";
          else
            cout << mn << " ";
        }
      } else
        cout << "NO";
    }
    cout << "\n";
  }
}