#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  int n;
  cin >> n;
  vector<int> vals(n);
  for (int& v : vals) cin >> v;
  vector<int> dp(2 * n, n + 1), pre(2 * n, -1);
  dp[0] = -n;
  dp[1] = -n;
  for (int i = 1; i < n; ++i) {
    int ind = 2 * i;
    for (int x = 0; x <= 1; ++x) {
      int vi = vals[i] * (2 * x - 1);
      for (int y = 0; y <= 1; ++y) {
        int vj = vals[i - 1] * (2 * y - 1);
        pair<int, int> pr = {dp[ind - 2 + y], vj};
        for (int s = 0; s <= 1; ++s, swap(pr.first, pr.second)) {
          if (pr.first <= vi && pr.second < dp[ind + x]) {
            dp[ind + x] = pr.second;
            pre[ind + x] = ind - 2 + y;
          }
        }
      }
    }
  }
  if (dp[2 * n - 1] <= n) {
    cout << "YES\n";
    int ind = 2 * n - 1;
    for (int i = n - 1; i >= 0; --i) {
      if ((ind % 2) == 0) vals[i] *= -1;
      ind = pre[ind];
    }
    for (int v : vals) cout << v << ' ';
    cout << '\n';
  } else {
    cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (int ti = 0; ti < t; ++ti) solve();
}

### 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;
using ll = long long;
void solve() {
  int n;
  cin >> n;
  vector<int> vals(n);
  for (int& v : vals) cin >> v;
  vector<int> dp(2 * n, n + 1), pre(2 * n, -1);
  dp[0] = -n;
  dp[1] = -n;
  for (int i = 1; i < n; ++i) {
    int ind = 2 * i;
    for (int x = 0; x <= 1; ++x) {
      int vi = vals[i] * (2 * x - 1);
      for (int y = 0; y <= 1; ++y) {
        int vj = vals[i - 1] * (2 * y - 1);
        pair<int, int> pr = {dp[ind - 2 + y], vj};
        for (int s = 0; s <= 1; ++s, swap(pr.first, pr.second)) {
          if (pr.first <= vi && pr.second < dp[ind + x]) {
            dp[ind + x] = pr.second;
            pre[ind + x] = ind - 2 + y;
          }
        }
      }
    }
  }
  if (dp[2 * n - 1] <= n) {
    cout << "YES\n";
    int ind = 2 * n - 1;
    for (int i = n - 1; i >= 0; --i) {
      if ((ind % 2) == 0) vals[i] *= -1;
      ind = pre[ind];
    }
    for (int v : vals) cout << v << ' ';
    cout << '\n';
  } else {
    cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (int ti = 0; ti < t; ++ti) solve();
}

```