#include <bits/stdc++.h>
using namespace std;
long long int t, n, a[1000007], dp[1000007][2][2];
pair<long long int, long long int> p[1000007][2][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long int i = 0; i < n; ++i) cin >> a[i];
    for (long long int i = 0; i < n; ++i)
      dp[i][0][0] = dp[i][0][1] = dp[i][1][1] = dp[i][1][0] =
          0x3f3f3f3f3f3f3f3f;
    dp[0][0][0] = dp[0][0][1] = -0x3f3f3f3f3f3f3f3f;
    for (long long int i = 0; i < n - 1; ++i) {
      for (long long int j = 0; j < 2; ++j) {
        for (long long int k = 0; k < 2; ++k) {
          if (dp[i][j][k] == 0x3f3f3f3f3f3f3f3f) continue;
          long long int x = a[i] * (k ? 1 : -1), y = dp[i][j][k];
          if (j) swap(x, y);
          for (long long int nxt = 0; nxt < 2; ++nxt) {
            long long int z = a[i + 1] * (nxt ? 1 : -1);
            if (z > x) {
              if (y < dp[i + 1][0][nxt]) {
                dp[i + 1][0][nxt] = y;
                p[i + 1][0][nxt] = {j, k};
              }
            } else if (z > y) {
              if (x < dp[i + 1][1][nxt]) {
                dp[i + 1][1][nxt] = x;
                p[i + 1][1][nxt] = {j, k};
              }
            }
          }
        }
      }
    }
    bool ok = 1;
    for (long long int i = 0; i < 2; ++i)
      for (long long int j = 0; j < 2; ++j)
        if (dp[n - 1][i][j] != 0x3f3f3f3f3f3f3f3f) ok = 0;
    if (ok) {
      cout << "NO\n";
      continue;
    }
    vector<long long int> ans;
    pair<long long int, long long int> x;
    for (long long int i = 0; i < 2; ++i)
      for (long long int j = 0; j < 2; ++j)
        if (dp[n - 1][i][j] != 0x3f3f3f3f3f3f3f3f) x = {i, j};
    for (long long int i = n - 1; i >= 0; --i) {
      ans.push_back(x.second ? a[i] : -a[i]);
      if (i > 0) x = p[i][x.first][x.second];
    }
    cout << "YES\n";
    reverse(ans.begin(), ans.end());
    for (long long int i : ans) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long int t, n, a[1000007], dp[1000007][2][2];
pair<long long int, long long int> p[1000007][2][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long int i = 0; i < n; ++i) cin >> a[i];
    for (long long int i = 0; i < n; ++i)
      dp[i][0][0] = dp[i][0][1] = dp[i][1][1] = dp[i][1][0] =
          0x3f3f3f3f3f3f3f3f;
    dp[0][0][0] = dp[0][0][1] = -0x3f3f3f3f3f3f3f3f;
    for (long long int i = 0; i < n - 1; ++i) {
      for (long long int j = 0; j < 2; ++j) {
        for (long long int k = 0; k < 2; ++k) {
          if (dp[i][j][k] == 0x3f3f3f3f3f3f3f3f) continue;
          long long int x = a[i] * (k ? 1 : -1), y = dp[i][j][k];
          if (j) swap(x, y);
          for (long long int nxt = 0; nxt < 2; ++nxt) {
            long long int z = a[i + 1] * (nxt ? 1 : -1);
            if (z > x) {
              if (y < dp[i + 1][0][nxt]) {
                dp[i + 1][0][nxt] = y;
                p[i + 1][0][nxt] = {j, k};
              }
            } else if (z > y) {
              if (x < dp[i + 1][1][nxt]) {
                dp[i + 1][1][nxt] = x;
                p[i + 1][1][nxt] = {j, k};
              }
            }
          }
        }
      }
    }
    bool ok = 1;
    for (long long int i = 0; i < 2; ++i)
      for (long long int j = 0; j < 2; ++j)
        if (dp[n - 1][i][j] != 0x3f3f3f3f3f3f3f3f) ok = 0;
    if (ok) {
      cout << "NO\n";
      continue;
    }
    vector<long long int> ans;
    pair<long long int, long long int> x;
    for (long long int i = 0; i < 2; ++i)
      for (long long int j = 0; j < 2; ++j)
        if (dp[n - 1][i][j] != 0x3f3f3f3f3f3f3f3f) x = {i, j};
    for (long long int i = n - 1; i >= 0; --i) {
      ans.push_back(x.second ? a[i] : -a[i]);
      if (i > 0) x = p[i][x.first][x.second];
    }
    cout << "YES\n";
    reverse(ans.begin(), ans.end());
    for (long long int i : ans) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}

```