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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long double PI = acos(-1.0);
bool AreSame(long double a, long double b) { return fabs(a - b) < 1e-9; }
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int ceil2(long long int a, long long int b) {
  long long int c = a / b;
  if (a % b) c++;
  return c;
}
bool isPalindrome(string s) {
  string t = s;
  reverse(t.begin(), t.end());
  return s == t;
}
void solve() {
  long long int x, y, z;
  cin >> x >> y >> z;
  map<long long int, long long int> mm;
  vector<long long int> v;
  v.push_back(x);
  v.push_back(y);
  v.push_back(z);
  mm[x]++;
  mm[y]++;
  mm[z]++;
  sort(v.begin(), v.end());
  if (mm[v[2]] == 1) {
    cout << "NO"
         << "\n";
    return;
  }
  cout << "YES"
       << "\n";
  if (mm[v[2]] == 3) {
    cout << v[2] << " " << v[2] << " " << v[2] << "\n";
    return;
  } else {
    if (x == v[2] && y == v[2]) {
      cout << v[2] << " " << z << " " << z << "\n";
    } else if (x == v[2] && z == v[2]) {
      cout << y << " " << v[2] << " " << y << "\n";
    } else if (z == v[2] && y == v[2]) {
      cout << x << " " << x << " " << v[2] << "\n";
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}