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() {
  int x, y, z;
  cin >> x >> y >> z;
  std::vector<int> v;
  v.push_back(x);
  v.push_back(y);
  v.push_back(z);
  set<int> st;
  st.insert(x);
  st.insert(y);
  st.insert(z);
  if (st.size() == 3) {
    cout << "NO" << endl;
  } else if (st.size() == 1) {
    cout << "YES" << endl;
    cout << x << " " << y << " " << z << endl;
  } else {
    int mx = *max_element(st.begin(), st.end());
    int cnt = count(v.begin(), v.end(), mx);
    if (cnt == 2) {
      cout << "YES" << endl;
      for (auto it : st) {
        if (it != mx) {
          cout << it << " " << it << " ";
          break;
        }
      }
      cout << mx << endl;
    } else
      cout << "NO" << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}