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;
using std::ios;
constexpr int MOD = 1e9 + 7;
constexpr int mod = 998244353;
long long hcf(long long a, long long b) {
  while (b) {
    long long r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    int a, b, c;
    cin >> a >> b >> c;
    if (a == b || b == c || c == a) {
      if (a == b && b < c)
        cout << "NO\n";
      else if (c == b && b < a)
        cout << "NO\n";
      else if (a == c && c < b)
        cout << "NO\n";
      else {
        cout << "YES\n";
        cout << max({a, b, c}) << ' ' << min({a, b, c}) << ' '
             << max(min({a, b, c}) - 1, 1) << '\n';
      }
    } else
      cout << "NO\n";
  }
}