#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 2;
const int INF = 1e9;
int dp[maxn][2];
int pre[maxn][2];
int p[maxn];
int a[maxn];
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 2; j++) {
      dp[i][j] = INF;
    }
  }
  dp[0][0] = -INF;
  dp[0][1] = -INF;
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0; j < 2; j++) {
      if (dp[i][j] != INF) {
        int x = j ? -p[i] : p[i];
        int y = dp[i][j];
        if (x < y) swap(x, y);
        for (int k = 0; k < 2; k++) {
          int z = k ? -p[i + 1] : p[i + 1];
          if (z > x) {
            if (dp[i + 1][k] > y) {
              dp[i + 1][k] = y;
              pre[i + 1][k] = j;
            }
          } else if (z > y) {
            if (dp[i + 1][k] > x) {
              dp[i + 1][k] = x;
              pre[i + 1][k] = j;
            }
          }
        }
      }
    }
  }
  for (int j = 0; j < 2; j++) {
    if (dp[n - 1][j] != INF) {
      for (int i = n - 1; i >= 0; i--) {
        a[i] = j ? -p[i] : p[i];
        j = pre[i][j];
      }
      cout << "YES\n";
      for (int i = 0; i < n; i++) {
        cout << a[i] << ' ';
      }
      cout << '\n';
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Please formulate 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;
const int maxn = 1e6 + 2;
const int INF = 1e9;
int dp[maxn][2];
int pre[maxn][2];
int p[maxn];
int a[maxn];
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 2; j++) {
      dp[i][j] = INF;
    }
  }
  dp[0][0] = -INF;
  dp[0][1] = -INF;
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0; j < 2; j++) {
      if (dp[i][j] != INF) {
        int x = j ? -p[i] : p[i];
        int y = dp[i][j];
        if (x < y) swap(x, y);
        for (int k = 0; k < 2; k++) {
          int z = k ? -p[i + 1] : p[i + 1];
          if (z > x) {
            if (dp[i + 1][k] > y) {
              dp[i + 1][k] = y;
              pre[i + 1][k] = j;
            }
          } else if (z > y) {
            if (dp[i + 1][k] > x) {
              dp[i + 1][k] = x;
              pre[i + 1][k] = j;
            }
          }
        }
      }
    }
  }
  for (int j = 0; j < 2; j++) {
    if (dp[n - 1][j] != INF) {
      for (int i = n - 1; i >= 0; i--) {
        a[i] = j ? -p[i] : p[i];
        j = pre[i][j];
      }
      cout << "YES\n";
      for (int i = 0; i < n; i++) {
        cout << a[i] << ' ';
      }
      cout << '\n';
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}

```