#include <bits/stdc++.h>
using namespace std;
int n;
int p[1000005];
int f[1000005][2][2];
tuple<int, int, int> pre[1000005][2][2];
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 0; i <= n; i++)
    for (int k1 = 0; k1 <= 1; k1++)
      for (int k2 = 0; k2 <= 1; k2++) f[i][k1][k2] = 1e9;
  f[1][0][0] = f[1][1][0] = -1e9;
  for (int i = 1; i < n; i++) {
    for (int k1 = 0; k1 <= 1; k1++) {
      for (int k2 = 0; k2 <= 1; k2++) {
        int x, y, t = f[i][k1][k2];
        x = (2 * k1 - 1) * p[i], y = t;
        if (k2) swap(x, y);
        auto nw = make_tuple(i, k1, k2);
        for (int sgn = 0; sgn <= 1; sgn++) {
          int z = (2 * sgn - 1) * p[i + 1];
          if (z < y) continue;
          if (z < x && f[i + 1][sgn][1] > x)
            f[i + 1][sgn][1] = x, pre[i + 1][sgn][1] = nw;
          if (z >= x && f[i + 1][sgn][0] > y)
            f[i + 1][sgn][0] = y, pre[i + 1][sgn][0] = nw;
        }
      }
    }
  }
  for (int k1 = 0; k1 <= 1; k1++) {
    for (int k2 = 0; k2 <= 1; k2++) {
      if (f[n][k1][k2] != 1e9) {
        vector<int> ans;
        int i = n;
        while (i) {
          ans.push_back((2 * k1 - 1) * p[i]);
          tie(i, k1, k2) = pre[i][k1][k2];
        }
        reverse(ans.begin(), ans.end());
        cout << "YES\n";
        for (auto x : ans) cout << x << ' ';
        cout << '\n';
        return;
      }
    }
  }
  cout << "NO\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  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 n;
int p[1000005];
int f[1000005][2][2];
tuple<int, int, int> pre[1000005][2][2];
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 0; i <= n; i++)
    for (int k1 = 0; k1 <= 1; k1++)
      for (int k2 = 0; k2 <= 1; k2++) f[i][k1][k2] = 1e9;
  f[1][0][0] = f[1][1][0] = -1e9;
  for (int i = 1; i < n; i++) {
    for (int k1 = 0; k1 <= 1; k1++) {
      for (int k2 = 0; k2 <= 1; k2++) {
        int x, y, t = f[i][k1][k2];
        x = (2 * k1 - 1) * p[i], y = t;
        if (k2) swap(x, y);
        auto nw = make_tuple(i, k1, k2);
        for (int sgn = 0; sgn <= 1; sgn++) {
          int z = (2 * sgn - 1) * p[i + 1];
          if (z < y) continue;
          if (z < x && f[i + 1][sgn][1] > x)
            f[i + 1][sgn][1] = x, pre[i + 1][sgn][1] = nw;
          if (z >= x && f[i + 1][sgn][0] > y)
            f[i + 1][sgn][0] = y, pre[i + 1][sgn][0] = nw;
        }
      }
    }
  }
  for (int k1 = 0; k1 <= 1; k1++) {
    for (int k2 = 0; k2 <= 1; k2++) {
      if (f[n][k1][k2] != 1e9) {
        vector<int> ans;
        int i = n;
        while (i) {
          ans.push_back((2 * k1 - 1) * p[i]);
          tie(i, k1, k2) = pre[i][k1][k2];
        }
        reverse(ans.begin(), ans.end());
        cout << "YES\n";
        for (auto x : ans) cout << x << ' ';
        cout << '\n';
        return;
      }
    }
  }
  cout << "NO\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

```