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;
const int mod = 1000000007;
bool sortFunction(string a, string b) {
  long long l1 = a.size(), l2 = b.size();
  if (l1 == l2) return (a < b);
  return l1 < l2;
}
unsigned long long power(unsigned long long x, unsigned long long y) {
  unsigned long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (ans * x);
    y >>= 1;
    x = (x * x);
  }
  return ans;
}
double nthRoot(int A, int N) {
  double pre = rand() % 10, eps = 1e-3;
  double del = INT_MAX;
  double xK;
  while (del > eps) {
    xK = ((N - 1.0) * pre + (double)A / pow(pre, N - 1)) / (double)N;
    del = abs(xK - pre);
    pre = xK;
  }
  return xK;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long a[3];
    for (long long i = 0; i < 3; i++) cin >> a[i];
    long long ct = count(a, a + 3, (*max_element(a, a + 3)));
    if (ct < 2)
      cout << "NO"
           << "\n";
    else {
      cout << "YES"
           << "\n";
      cout << *max_element(a, a + 3) << " " << *min_element(a, a + 3) << " "
           << "1"
           << "\n";
    }
  }
}