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() {
  map<int, int> p;
  int x, y, z;
  cin >> x >> y >> z;
  p[x] = 1;
  if (p.find(y) != p.end()) {
    p[y]++;
  } else
    p[y] = 1;
  if (p.find(z) != p.end()) {
    p[z]++;
  } else
    p[z] = 1;
  int a, b, c;
  if (p.size() == 3)
    cout << "NO\n";
  else if (p.size() == 1)
    cout << "YES\n" << x << " " << x << " " << x << "\n";
  if (p.size() == 2) {
    auto it = p.begin();
    auto it1 = it++;
    if ((it->first > it1->first) && (it->second > it1->second)) {
      cout << "YES\n1 " << it->first << " " << it1->first << endl;
    } else if ((it->first < it1->first) && (it->second < it1->second)) {
      cout << "YES\n1 " << it->first << " " << it1->first << endl;
    } else
      cout << "NO\n";
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}