#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int main() {
  std::ios_base::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    vector<vector<int>> f(n + 1, vector<int>(4, inf));
    vector<vector<int>> pre(n + 1, vector<int>(4, -1));
    f[1][0] = f[1][1] = -inf;
    for (int i = 2; i <= n; i++) {
      f[i] = vector<int>(4, inf);
      vector<tuple<int, int, int>> b;
      for (int j = 0; j < 4; j++) {
        int x = (j % 2 ? -1 : 1) * a[i - 1];
        if (f[i - 1][j] != inf) {
          int y = f[i - 1][j];
          if (x < y) {
            swap(x, y);
          }
          b.emplace_back(j, x, y);
        }
      }
      for (int j = 0; j < 4; j++) {
        int u = (j % 2 ? -1 : 1) * a[i];
        for (auto [p, x, y] : b) {
          if (j < 2 && u > x) {
            if (y < f[i][j]) {
              f[i][j] = y;
              pre[i][j] = p;
            }
          }
          if (j >= 2 && u > y && u < x) {
            if (x < f[i][j]) {
              f[i][j] = x;
              pre[i][j] = p;
            }
          }
        }
      }
    }
    bool ok = false;
    vector<int> ans(n + 1);
    for (int j = 0; j < 4; j++) {
      if (f[n][j] < inf) {
        int p = j;
        ok = true;
        for (int i = n; i >= 1; i--) {
          ans[i] = (p % 2 ? -1 : 1) * a[i];
          p = pre[i][p];
        }
        break;
      }
    }
    cout << (ok ? "YES" : "NO") << endl;
    if (ok) {
      for (int i = 1; i <= n; i++) {
        cout << ans[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 inf = 1e9;
int main() {
  std::ios_base::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    vector<vector<int>> f(n + 1, vector<int>(4, inf));
    vector<vector<int>> pre(n + 1, vector<int>(4, -1));
    f[1][0] = f[1][1] = -inf;
    for (int i = 2; i <= n; i++) {
      f[i] = vector<int>(4, inf);
      vector<tuple<int, int, int>> b;
      for (int j = 0; j < 4; j++) {
        int x = (j % 2 ? -1 : 1) * a[i - 1];
        if (f[i - 1][j] != inf) {
          int y = f[i - 1][j];
          if (x < y) {
            swap(x, y);
          }
          b.emplace_back(j, x, y);
        }
      }
      for (int j = 0; j < 4; j++) {
        int u = (j % 2 ? -1 : 1) * a[i];
        for (auto [p, x, y] : b) {
          if (j < 2 && u > x) {
            if (y < f[i][j]) {
              f[i][j] = y;
              pre[i][j] = p;
            }
          }
          if (j >= 2 && u > y && u < x) {
            if (x < f[i][j]) {
              f[i][j] = x;
              pre[i][j] = p;
            }
          }
        }
      }
    }
    bool ok = false;
    vector<int> ans(n + 1);
    for (int j = 0; j < 4; j++) {
      if (f[n][j] < inf) {
        int p = j;
        ok = true;
        for (int i = n; i >= 1; i--) {
          ans[i] = (p % 2 ? -1 : 1) * a[i];
          p = pre[i][p];
        }
        break;
      }
    }
    cout << (ok ? "YES" : "NO") << endl;
    if (ok) {
      for (int i = 1; i <= n; i++) {
        cout << ans[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}

```