#include <bits/stdc++.h>
using namespace std;
long long a[1000005], dp[1000005][2];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    a[0] = 1e9, dp[0][0] = -1e9;
    for (long long i = 1; i <= n; i++) dp[i][0] = dp[i][1] = 1e9;
    for (long long i = 1; i <= n; i++) {
      long long x = -a[i - 1], y = dp[i - 1][0];
      if (a[i] > x) dp[i][1] = min(dp[i][1], y);
      if (a[i] > y) dp[i][1] = min(dp[i][1], x);
      if (-a[i] > x) dp[i][0] = min(dp[i][0], y);
      if (-a[i] > y) dp[i][0] = min(dp[i][0], x);
      x = a[i - 1], y = dp[i - 1][1];
      if (a[i] > x) dp[i][1] = min(dp[i][1], y);
      if (a[i] > y) dp[i][1] = min(dp[i][1], x);
      if (-a[i] > x) dp[i][0] = min(dp[i][0], y);
      if (-a[i] > y) dp[i][0] = min(dp[i][0], x);
    }
    if (dp[n][1] <= n || dp[n][0] <= n) {
      cout << "YES\n";
      long long now = 0;
      if (dp[n][1] <= n) now = 1;
      for (long long i = n; i >= 1; i--) {
        if (now == 1) {
          long long x = -a[i - 1], y = dp[i - 1][0];
          if (a[i] > x && dp[i][1] == y) now = 0;
          if (a[i] > y && dp[i][1] == x) now = 0;
          x = a[i - 1], y = dp[i - 1][1];
          if (a[i] > x && dp[i][1] == y) now = 1;
          if (a[i] > y && dp[i][1] == x) now = 1;
        } else {
          long long x = -a[i - 1], y = dp[i - 1][0];
          if (-a[i] > x && dp[i][0] == y) now = 0;
          if (-a[i] > y && dp[i][0] == x) now = 0;
          x = a[i - 1], y = dp[i - 1][1];
          if (-a[i] > x && dp[i][0] == y) now = 1;
          if (-a[i] > y && dp[i][0] == x) now = 1;
          a[i] = -a[i];
        }
      }
      for (long long i = 1; i <= n; i++) cout << a[i] << " ";
      cout << "\n";
    } else
      cout << "NO\n";
  }
  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;
long long a[1000005], dp[1000005][2];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    a[0] = 1e9, dp[0][0] = -1e9;
    for (long long i = 1; i <= n; i++) dp[i][0] = dp[i][1] = 1e9;
    for (long long i = 1; i <= n; i++) {
      long long x = -a[i - 1], y = dp[i - 1][0];
      if (a[i] > x) dp[i][1] = min(dp[i][1], y);
      if (a[i] > y) dp[i][1] = min(dp[i][1], x);
      if (-a[i] > x) dp[i][0] = min(dp[i][0], y);
      if (-a[i] > y) dp[i][0] = min(dp[i][0], x);
      x = a[i - 1], y = dp[i - 1][1];
      if (a[i] > x) dp[i][1] = min(dp[i][1], y);
      if (a[i] > y) dp[i][1] = min(dp[i][1], x);
      if (-a[i] > x) dp[i][0] = min(dp[i][0], y);
      if (-a[i] > y) dp[i][0] = min(dp[i][0], x);
    }
    if (dp[n][1] <= n || dp[n][0] <= n) {
      cout << "YES\n";
      long long now = 0;
      if (dp[n][1] <= n) now = 1;
      for (long long i = n; i >= 1; i--) {
        if (now == 1) {
          long long x = -a[i - 1], y = dp[i - 1][0];
          if (a[i] > x && dp[i][1] == y) now = 0;
          if (a[i] > y && dp[i][1] == x) now = 0;
          x = a[i - 1], y = dp[i - 1][1];
          if (a[i] > x && dp[i][1] == y) now = 1;
          if (a[i] > y && dp[i][1] == x) now = 1;
        } else {
          long long x = -a[i - 1], y = dp[i - 1][0];
          if (-a[i] > x && dp[i][0] == y) now = 0;
          if (-a[i] > y && dp[i][0] == x) now = 0;
          x = a[i - 1], y = dp[i - 1][1];
          if (-a[i] > x && dp[i][0] == y) now = 1;
          if (-a[i] > y && dp[i][0] == x) now = 1;
          a[i] = -a[i];
        }
      }
      for (long long i = 1; i <= n; i++) cout << a[i] << " ";
      cout << "\n";
    } else
      cout << "NO\n";
  }
  return 0;
}

```