#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7, inf = 1e9;
int n, dp[N][2], w[N][2], lst[N][2], a[N], ns[N];
void Main() {
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i], w[i][0] = a[i], w[i][1] = -a[i];
  for (int i = (0); i <= (n); i++) dp[i][0] = dp[i][1] = inf;
  w[0][0] = w[0][1] = -inf;
  dp[0][0] = dp[0][1] = -inf;
  for (int i = (1); i <= (n); i++) {
    for (int a = (0); a <= (1); a++)
      for (int b = (0); b <= (1); b++)
        if (w[i][a] > w[i - 1][b] && dp[i][a] > dp[i - 1][b])
          dp[i][a] = dp[i - 1][b], lst[i][a] = b;
    for (int a = (0); a <= (1); a++)
      for (int b = (0); b <= (1); b++)
        if (w[i][a] > dp[i - 1][b] && dp[i][a] > w[i - 1][b])
          dp[i][a] = w[i - 1][b], lst[i][a] = b;
  }
  if (dp[n][0] == inf && dp[n][1] == inf)
    cout << "NO\n";
  else {
    cout << "YES\n";
    int o = dp[n][0] == inf ? 1 : 0;
    for (int i = (n); i >= (1); i--) ns[i] = w[i][o], o = lst[i][o];
    for (int i = (1); i <= (n); i++) cout << ns[i] << ' ';
    cout << '\n';
  }
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) Main();
  return 0;
}

### Prompt

Please create a solution in cpp 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 N = 1e6 + 7, inf = 1e9;
int n, dp[N][2], w[N][2], lst[N][2], a[N], ns[N];
void Main() {
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i], w[i][0] = a[i], w[i][1] = -a[i];
  for (int i = (0); i <= (n); i++) dp[i][0] = dp[i][1] = inf;
  w[0][0] = w[0][1] = -inf;
  dp[0][0] = dp[0][1] = -inf;
  for (int i = (1); i <= (n); i++) {
    for (int a = (0); a <= (1); a++)
      for (int b = (0); b <= (1); b++)
        if (w[i][a] > w[i - 1][b] && dp[i][a] > dp[i - 1][b])
          dp[i][a] = dp[i - 1][b], lst[i][a] = b;
    for (int a = (0); a <= (1); a++)
      for (int b = (0); b <= (1); b++)
        if (w[i][a] > dp[i - 1][b] && dp[i][a] > w[i - 1][b])
          dp[i][a] = w[i - 1][b], lst[i][a] = b;
  }
  if (dp[n][0] == inf && dp[n][1] == inf)
    cout << "NO\n";
  else {
    cout << "YES\n";
    int o = dp[n][0] == inf ? 1 : 0;
    for (int i = (n); i >= (1); i--) ns[i] = w[i][o], o = lst[i][o];
    for (int i = (1); i <= (n); i++) cout << ns[i] << ' ';
    cout << '\n';
  }
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) Main();
  return 0;
}

```