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;
bool checkPrime(int n);
vector<int> primeFactors(int n);
void solve() {
  int x, y, z;
  cin >> x >> y >> z;
  int a = min(x, y), b = min(x, z), c = min(y, z);
  if (x == max(a, b) && y == max(a, c) & z == max(b, c))
    cout << "YES\n" << a << " " << b << " " << c << endl;
  else
    cout << "NO\n";
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
bool checkPrime(int n) {
  if (n % 2 == 0) return false;
  for (int i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
vector<int> primeFactors(int n) {
  vector<int> v;
  while (n % 2 == 0) {
    v.push_back(2);
    ;
    n /= 2;
  }
  for (int i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      v.push_back(i);
      ;
      n /= i;
    }
  }
  if (n > 2) v.push_back(n);
  ;
  return v;
}