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>
unsigned long long int a, b, c;
unsigned long long int t, x, y, z, i;
void word(void);
int main() {
  scanf("%llu", &t);
  for (i = 1; i <= t; i++) {
    scanf("%llu %llu %llu", &x, &y, &z);
    if ((x == y) || (y == z) || (x == z)) {
      if (y == z && y > x) {
        a = z;
        b = x;
        c = x - 1;
        word();
      } else if (y == x && y > z) {
        a = y;
        b = z;
        c = z - 1;
        word();
      } else if (z == x && z > y) {
        a = z;
        b = y;
        c = y - 1;
        word();
      } else if (y == x && x == z && y == z) {
        a = x;
        b = y;
        c = z;
        word();
      } else
        printf("NO");
    } else
      printf("NO");
    printf("\n");
  }
}
void word(void) {
  if (c > 0) printf("YES\n%llu %llu %llu", a, b, c);
  if (c == 0) printf("YES\n%llu %llu %llu", a, b, c + 1);
}