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>
const int mx = 1e6 + 5;
const int inf = 0x3f3f3f3f;
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  scanf("%d", &t);
  while (t--) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (x == y && y == z) {
      printf("YES\n%d %d %d\n", x, y, z);
      continue;
    }
    if (x == y && y > z) {
      printf("YES\n%d %d %d\n", x, min(x, min(y, z)), min(x, min(y, z)));
      continue;
    }
    if (x < y && y == z) {
      printf("YES\n%d %d %d\n", min(x, min(y, z)), min(x, min(y, z)), y);
      continue;
    }
    if (x == z && x > y) {
      printf("YES\n%d %d %d\n", min(x, min(y, z)), x, min(x, min(y, z)));
      continue;
    }
    puts("NO");
  }
  return 0;
}