You are given a permutation p consisting of n integers 1, 2, ..., n (a permutation is an array where each element from 1 to n occurs exactly once).

Let's call an array a bipartite if the following undirected graph is bipartite:

  * the graph consists of n vertices; 
  * two vertices i and j are connected by an edge if i < j and a_i > a_j. 



Your task is to find a bipartite array of integers a of size n, such that a_i = p_i or a_i = -p_i, or report that no such array exists. If there are multiple answers, print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^6) — the size of the permutation.

The second line contains n integers p_1, p_2, ..., p_n.

The sum of n over all test cases doesn't exceed 10^6.

Output

For each test case, print the answer in the following format. If such an array a does not exist, print "NO" in a single line. Otherwise, print "YES" in the first line and n integers — array a in the second line.

Example

Input


4
3
1 2 3
6
1 3 2 6 5 4
4
4 1 3 2
8
3 2 1 6 7 8 5 4


Output


YES
1 2 3
NO
YES
-4 -1 -3 -2
YES
-3 -2 1 6 7 -8 -5 -4
4
3
1 2 3
6
1 3 2 6 5 4
4
4 1 3 2
8
3 2 1 6 7 8 5 4

YES
1 2 3
NO
YES
-4 -1 -3 2
YES
-3 -2 1 6 7 -8 -5 4

#include <bits/stdc++.h>
using namespace std;
long long a[1000005];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    long long x = -1e10, y = -1e9, flag = 1;
    for (long long i = 1; i <= n; i++) {
      if (-a[i] > y)
        y = -a[i], a[i] = -a[i];
      else if (-a[i] > x)
        x = -a[i], a[i] = -a[i];
      else if (a[i] > y)
        y = a[i];
      else if (a[i] > x)
        x = a[i];
      else
        flag = 0;
    }
    if (flag) {
      cout << "YES\n";
      for (long long i = 1; i <= n; i++) cout << a[i] << " ";
      cout << "\n";
    } else
      cout << "NO\n";
  }
  return 0;
}
