#include <bits/stdc++.h>
using namespace std;
constexpr int kN = 1e6 + 10;
int dp[kN][2][2];
pair<int, int> prefix[kN][2][2];
void Solve() {
  constexpr int kInf = 1e9;
  int n;
  cin >> n;
  vector<int> p(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++) {
      for (int k = 0; k < 2; k++) dp[i][j][k] = kInf;
    }
  }
  dp[0][0][0] = dp[0][0][1] = -kInf;
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0; j < 2; j++) {
      for (int k = 0; k < 2; k++)
        if (dp[i][j][k] != kInf) {
          int x = k == 0 ? p[i] : -p[i];
          int y = dp[i][j][k];
          if (x < y) swap(x, y);
          for (int o = 0; o < 2; o++) {
            int new_x = o == 0 ? p[i + 1] : -p[i + 1];
            if (new_x > x) {
              if (dp[i + 1][0][o] > y) {
                dp[i + 1][0][o] = y;
                prefix[i + 1][0][o] = make_pair(j, k);
              }
            } else if (new_x > y) {
              if (dp[i + 1][1][o] > x) {
                dp[i + 1][1][o] = x;
                prefix[i + 1][1][o] = make_pair(j, k);
              }
            }
          }
        }
    }
  }
  int nj = -1, nk = -1;
  for (int j = 0; j < 2; j++) {
    for (int k = 0; k < 2; k++) {
      if (dp[n - 1][j][k] != kInf) nj = j, nk = k;
    }
  }
  if (nj == -1 && nk == -1) {
    cout << "NO\n";
    return;
  }
  vector<int> a(n);
  for (int i = n - 1; i >= 0; i--) {
    a[i] = nk == 0 ? p[i] : -p[i];
    tie(nj, nk) = prefix[i][nj][nk];
  }
  cout << "YES\n";
  for (auto x : a) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) Solve();
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
constexpr int kN = 1e6 + 10;
int dp[kN][2][2];
pair<int, int> prefix[kN][2][2];
void Solve() {
  constexpr int kInf = 1e9;
  int n;
  cin >> n;
  vector<int> p(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++) {
      for (int k = 0; k < 2; k++) dp[i][j][k] = kInf;
    }
  }
  dp[0][0][0] = dp[0][0][1] = -kInf;
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0; j < 2; j++) {
      for (int k = 0; k < 2; k++)
        if (dp[i][j][k] != kInf) {
          int x = k == 0 ? p[i] : -p[i];
          int y = dp[i][j][k];
          if (x < y) swap(x, y);
          for (int o = 0; o < 2; o++) {
            int new_x = o == 0 ? p[i + 1] : -p[i + 1];
            if (new_x > x) {
              if (dp[i + 1][0][o] > y) {
                dp[i + 1][0][o] = y;
                prefix[i + 1][0][o] = make_pair(j, k);
              }
            } else if (new_x > y) {
              if (dp[i + 1][1][o] > x) {
                dp[i + 1][1][o] = x;
                prefix[i + 1][1][o] = make_pair(j, k);
              }
            }
          }
        }
    }
  }
  int nj = -1, nk = -1;
  for (int j = 0; j < 2; j++) {
    for (int k = 0; k < 2; k++) {
      if (dp[n - 1][j][k] != kInf) nj = j, nk = k;
    }
  }
  if (nj == -1 && nk == -1) {
    cout << "NO\n";
    return;
  }
  vector<int> a(n);
  for (int i = n - 1; i >= 0; i--) {
    a[i] = nk == 0 ? p[i] : -p[i];
    tie(nj, nk) = prefix[i][nj][nk];
  }
  cout << "YES\n";
  for (auto x : a) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) Solve();
  return 0;
}

```