#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<vector<int>> a(n, vector<int>(2));
    vector<vector<pair<int, int>>> dp(n, vector<pair<int, int>>(2));
    for (int i = 0; i < n; i++) {
      cin >> a[i][1];
      a[i][0] = -a[i][1];
      if (i == 0) {
        dp[i][0] = dp[i][1] = {-n - 1, 0};
      } else {
        for (int j = 0; j < 2; j++) {
          dp[i][j] = {n + 1, 0};
          for (int k = 0; k < 2; k++) {
            if (a[i - 1][k] < a[i][j]) {
              dp[i][j] = min(dp[i][j], {dp[i - 1][k].first, k});
            }
            if (dp[i - 1][k].first < a[i][j]) {
              dp[i][j] = min(dp[i][j], {a[i - 1][k], k});
            }
          }
        }
      }
    }
    for (int cur = 0; cur < 2; cur++) {
      if (dp[n - 1][cur].first <= n) {
        cout << "YES\n";
        for (int i = n - 1; i >= 0; i--) {
          a[i][0] = a[i][cur];
          cur = dp[i][cur].second;
        }
        for (int i = 0; i < n; i++) {
          cout << a[i][0] << " \n"[i == n - 1];
        }
        goto hell;
      }
    }
    cout << "NO\n";
  hell:
    continue;
  }
}

### Prompt

Generate a cpp solution to the following problem:

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(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<vector<int>> a(n, vector<int>(2));
    vector<vector<pair<int, int>>> dp(n, vector<pair<int, int>>(2));
    for (int i = 0; i < n; i++) {
      cin >> a[i][1];
      a[i][0] = -a[i][1];
      if (i == 0) {
        dp[i][0] = dp[i][1] = {-n - 1, 0};
      } else {
        for (int j = 0; j < 2; j++) {
          dp[i][j] = {n + 1, 0};
          for (int k = 0; k < 2; k++) {
            if (a[i - 1][k] < a[i][j]) {
              dp[i][j] = min(dp[i][j], {dp[i - 1][k].first, k});
            }
            if (dp[i - 1][k].first < a[i][j]) {
              dp[i][j] = min(dp[i][j], {a[i - 1][k], k});
            }
          }
        }
      }
    }
    for (int cur = 0; cur < 2; cur++) {
      if (dp[n - 1][cur].first <= n) {
        cout << "YES\n";
        for (int i = n - 1; i >= 0; i--) {
          a[i][0] = a[i][cur];
          cur = dp[i][cur].second;
        }
        for (int i = 0; i < n; i++) {
          cout << a[i][0] << " \n"[i == n - 1];
        }
        goto hell;
      }
    }
    cout << "NO\n";
  hell:
    continue;
  }
}

```