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;
long long int mod = 998244353;
long long int ceil1(long long int n, long long int k) {
  if (n % k) return n / k + 1;
  return n / k;
}
long long int pow1(long long int n, long long int k) {
  long long int ans = 1;
  while (k != 0) {
    if (k % 2) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k = k / 2;
  }
  return ans;
}
bool ispal(string s) {
  long long int i = 0;
  long long int j = s.size() - 1;
  while (i < j) {
    if (s[i] != s[j]) return false;
    i++;
    j--;
  }
  return true;
}
long long int gcdd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcdd(b, a % b);
}
long long int fact(long long int n) {
  if (n == 1 || n == 0)
    return 1;
  else
    return (n * fact(n - 1)) % mod;
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool f(long long int a, long long int b, long long int c,
       vector<long long int> ar) {
  return ar[0] == max(a, b) && ar[1] == max(a, c) && ar[2] == max(b, c);
}
void solve() {
  vector<long long int> a(3);
  for (long long int i = 0; i < 3; i++) cin >> a[i];
  for (long long int i = 0; i < 3; i++) {
    for (long long int j = 0; j < 3; j++) {
      for (long long int k = 0; k < 3; k++) {
        if (f(a[i], a[j], a[k], a)) {
          cout << "YES" << endl;
          cout << a[i] << " " << a[j] << " " << a[k] << endl;
          return;
        }
      }
    }
  }
  cout << "NO" << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}