#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int &x : v) cin >> x;
    vector<vector<int>> dp(n, vector<int>(2, (int)1e9)),
        elozo(n, vector<int>(2, 0));
    dp[0] = {-(int)1e9, -(int)1e9};
    for (int i = 1; i < n; i++) {
      dp[i][0] = dp[i - 1][1];
      if (v[i] > dp[i - 1][1]) dp[i][0] = min(dp[i][0], -v[i - 1]);
      elozo[i][0] = 1;
      if (v[i] > dp[i - 1][0] && dp[i][0] > v[i - 1]) {
        dp[i][0] = v[i - 1];
        elozo[i][0] = 0;
      }
      if (v[i] > v[i - 1] && dp[i][0] > dp[i - 1][0]) {
        dp[i][0] = dp[i - 1][0];
        elozo[i][0] = 0;
      }
      if (-v[i] > dp[i - 1][0] && dp[i][1] > v[i - 1]) {
        dp[i][1] = v[i - 1];
        elozo[i][1] = 0;
      }
      if (-v[i] > dp[i - 1][1] && dp[i][1] > -v[i - 1]) {
        dp[i][1] = -v[i - 1];
        elozo[i][1] = 1;
      }
      if (-v[i] > -v[i - 1] && dp[i][1] > dp[i - 1][1]) {
        dp[i][1] = dp[i - 1][1];
        elozo[i][1] = 1;
      }
    }
    if (dp[n - 1][0] == (int)1e9) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    int x = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (x) v[i] = -v[i];
      x = elozo[i][x];
    }
    for (int &x : v) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int &x : v) cin >> x;
    vector<vector<int>> dp(n, vector<int>(2, (int)1e9)),
        elozo(n, vector<int>(2, 0));
    dp[0] = {-(int)1e9, -(int)1e9};
    for (int i = 1; i < n; i++) {
      dp[i][0] = dp[i - 1][1];
      if (v[i] > dp[i - 1][1]) dp[i][0] = min(dp[i][0], -v[i - 1]);
      elozo[i][0] = 1;
      if (v[i] > dp[i - 1][0] && dp[i][0] > v[i - 1]) {
        dp[i][0] = v[i - 1];
        elozo[i][0] = 0;
      }
      if (v[i] > v[i - 1] && dp[i][0] > dp[i - 1][0]) {
        dp[i][0] = dp[i - 1][0];
        elozo[i][0] = 0;
      }
      if (-v[i] > dp[i - 1][0] && dp[i][1] > v[i - 1]) {
        dp[i][1] = v[i - 1];
        elozo[i][1] = 0;
      }
      if (-v[i] > dp[i - 1][1] && dp[i][1] > -v[i - 1]) {
        dp[i][1] = -v[i - 1];
        elozo[i][1] = 1;
      }
      if (-v[i] > -v[i - 1] && dp[i][1] > dp[i - 1][1]) {
        dp[i][1] = dp[i - 1][1];
        elozo[i][1] = 1;
      }
    }
    if (dp[n - 1][0] == (int)1e9) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    int x = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (x) v[i] = -v[i];
      x = elozo[i][x];
    }
    for (int &x : v) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}

```