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;
void solve() {
  int n;
  cin >> n;
  int d[2] = {0, 0};
  int l[2] = {0, 0};
  int tp[2] = {n, n};
  vector<int> p(n);
  vector<int> s(n);
  bool fail = false;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    if (fail) continue;
    bool ok = false;
    for (int k = 0; k < 2; k++) {
      s[i] = k;
      if (l[k] == 0) {
        l[k] = p[i];
        d[k] = -1;
        ok = true;
        break;
      }
      if (d[k] == -1) {
        if (p[i] < l[k]) {
          l[k] = p[i];
        } else {
          d[k] = 1;
          tp[k] = i;
          l[k] = p[i];
        }
        ok = true;
        break;
      }
      if (p[i] > l[k]) {
        l[k] = p[i];
        ok = true;
        break;
      }
    }
    if (!ok) {
      cout << "NO" << endl;
      fail = true;
    }
  }
  if (fail) return;
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    if (i >= tp[s[i]]) {
      cout << p[i] << " ";
    } else {
      cout << -p[i] << " ";
    }
  }
  cout << endl;
}
int main() {
  int ncases;
  cin >> ncases;
  for (int i = 1; i <= ncases; i++) {
    solve();
  }
  return 0;
}
