#include <bits/stdc++.h>
using namespace std;
const int N = 1000010, INF = 0x3f3f3f3f;
int n;
int p[N], a[N];
int f[N][2], pre[N][2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _;
  cin >> _;
  while (_--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> p[i];
    f[1][0] = f[1][1] = -INF;
    for (int i = 2; i <= n; i++) {
      f[i][0] = f[i][1] = INF;
      if (p[i] > f[i - 1][0]) {
        if (f[i][0] > p[i - 1]) {
          f[i][0] = p[i - 1];
          pre[i][0] = 0;
        }
      }
      if (p[i] > p[i - 1]) {
        if (f[i][0] > f[i - 1][0]) {
          f[i][0] = f[i - 1][0];
          pre[i][0] = 0;
        }
      }
      if (p[i] > f[i - 1][1]) {
        if (f[i][0] > -p[i - 1]) {
          f[i][0] = -p[i - 1];
          pre[i][0] = 1;
        }
      }
      if (p[i] > -p[i - 1]) {
        if (f[i][0] > f[i - 1][1]) {
          f[i][0] = f[i - 1][1];
          pre[i][0] = 1;
        }
      }
      if (-p[i] > f[i - 1][0]) {
        if (f[i][1] > p[i - 1]) {
          f[i][1] = p[i - 1];
          pre[i][1] = 0;
        }
      }
      if (-p[i] > p[i - 1]) {
        if (f[i][1] > f[i - 1][0]) {
          f[i][1] = f[i - 1][0];
          pre[i][1] = 0;
        }
      }
      if (-p[i] > f[i - 1][1]) {
        if (f[i][1] > -p[i - 1]) {
          f[i][1] = -p[i - 1];
          pre[i][1] = 1;
        }
      }
      if (-p[i] > -p[i - 1]) {
        if (f[i][1] > f[i - 1][1]) {
          f[i][1] = f[i - 1][1];
          pre[i][1] = 1;
        }
      }
    }
    if (f[n][0] == INF && f[n][1] == INF)
      cout << "NO" << endl;
    else {
      int ed = f[n][0] != INF ? 0 : 1;
      for (int i = n; i >= 1; i--) {
        a[i] = ed == 0 ? p[i] : -p[i];
        ed = pre[i][ed];
      }
      cout << "YES" << endl;
      for (int i = 1; i <= n; i++) cout << a[i] << ' ';
      cout << endl;
    }
  }
  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;
const int N = 1000010, INF = 0x3f3f3f3f;
int n;
int p[N], a[N];
int f[N][2], pre[N][2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _;
  cin >> _;
  while (_--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> p[i];
    f[1][0] = f[1][1] = -INF;
    for (int i = 2; i <= n; i++) {
      f[i][0] = f[i][1] = INF;
      if (p[i] > f[i - 1][0]) {
        if (f[i][0] > p[i - 1]) {
          f[i][0] = p[i - 1];
          pre[i][0] = 0;
        }
      }
      if (p[i] > p[i - 1]) {
        if (f[i][0] > f[i - 1][0]) {
          f[i][0] = f[i - 1][0];
          pre[i][0] = 0;
        }
      }
      if (p[i] > f[i - 1][1]) {
        if (f[i][0] > -p[i - 1]) {
          f[i][0] = -p[i - 1];
          pre[i][0] = 1;
        }
      }
      if (p[i] > -p[i - 1]) {
        if (f[i][0] > f[i - 1][1]) {
          f[i][0] = f[i - 1][1];
          pre[i][0] = 1;
        }
      }
      if (-p[i] > f[i - 1][0]) {
        if (f[i][1] > p[i - 1]) {
          f[i][1] = p[i - 1];
          pre[i][1] = 0;
        }
      }
      if (-p[i] > p[i - 1]) {
        if (f[i][1] > f[i - 1][0]) {
          f[i][1] = f[i - 1][0];
          pre[i][1] = 0;
        }
      }
      if (-p[i] > f[i - 1][1]) {
        if (f[i][1] > -p[i - 1]) {
          f[i][1] = -p[i - 1];
          pre[i][1] = 1;
        }
      }
      if (-p[i] > -p[i - 1]) {
        if (f[i][1] > f[i - 1][1]) {
          f[i][1] = f[i - 1][1];
          pre[i][1] = 1;
        }
      }
    }
    if (f[n][0] == INF && f[n][1] == INF)
      cout << "NO" << endl;
    else {
      int ed = f[n][0] != INF ? 0 : 1;
      for (int i = n; i >= 1; i--) {
        a[i] = ed == 0 ? p[i] : -p[i];
        ed = pre[i][ed];
      }
      cout << "YES" << endl;
      for (int i = 1; i <= n; i++) cout << a[i] << ' ';
      cout << endl;
    }
  }
  return 0;
}

```