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;
int p[1000001];
int inc[1000001];
int decc[1000001];
int main() {
  int t, n;
  cin >> t;
outer:
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> p[i];
    }
    int last = -1000001;
    int i;
    for (i = 0; i < n; i++) {
      if (-p[i] > last) {
        inc[i] = 1;
        last = -p[i];
        continue;
      }
      if (p[i] > last) {
        inc[i] = 1;
        last = p[i];
        continue;
      }
      break;
    }
    for (; i < n; i++) inc[i] = 0;
    last = 1000001;
    decc[n] = 1;
    for (i = n - 1; i >= 0; i--) {
      if (p[i] < last) {
        decc[i] = 1;
        last = p[i];
        continue;
      }
      if (-p[i] < last) {
        decc[i] = 1;
        last = -p[i];
        continue;
      }
      break;
    }
    for (; i >= 0; i--) decc[i] = 0;
    int done = 0;
    for (i = 0; i < n; i++) {
      if (inc[i] && decc[i + 1]) {
        cout << "YES " << endl;
        last = -1000001;
        for (int j = 0; j <= i; j++) {
          if (-p[j] > last)
            last = -p[j];
          else
            last = p[j];
          cout << last << " ";
        }
        last = -1000001;
        for (int j = i + 1; j < n; j++) {
          if (-p[j] > last)
            last = -p[j];
          else
            last = p[j];
          cout << last << " ";
        }
        cout << endl;
        done = 1;
        break;
      }
    }
    if (!done) cout << "NO" << endl;
  }
}
