#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
const int MAXN = 1e6 + 1;
int dp[2][MAXN], p[2][MAXN], par[2][MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> p[0][i];
      p[1][i] = -p[0][i];
    }
    p[0][0] = p[1][0] = -1e9;
    dp[0][0] = dp[1][0] = -1e9;
    for (int i = 1; i <= n; i++) {
      dp[0][i] = dp[1][i] = 1e9;
      for (int j = 0; j < 2; j++) {
        if (dp[j][i - 1] == 1e9) continue;
        for (int k = 0; k < 2; k++) {
          if (dp[j][i - 1] < p[k][i] && p[j][i - 1] < dp[k][i]) {
            dp[k][i] = p[j][i - 1];
            par[k][i] = j;
          }
          if (p[j][i - 1] < p[k][i] && dp[j][i - 1] < dp[k][i]) {
            dp[k][i] = dp[j][i - 1];
            par[k][i] = j;
          }
        }
      }
    }
    if (dp[0][n] == 1e9 && dp[1][n] == 1e9)
      cout << "NO" << '\n';
    else {
      int c = dp[0][n] == 1e9;
      vector<int> ans;
      for (int i = n; i >= 1; i--) {
        ans.push_back(p[c][i]);
        c = par[c][i];
      }
      cout << "YES" << '\n';
      reverse(ans.begin(), ans.end());
      for (int x : ans) cout << x << " ";
      cout << '\n';
    }
  }
}

### 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 gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
const int MAXN = 1e6 + 1;
int dp[2][MAXN], p[2][MAXN], par[2][MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> p[0][i];
      p[1][i] = -p[0][i];
    }
    p[0][0] = p[1][0] = -1e9;
    dp[0][0] = dp[1][0] = -1e9;
    for (int i = 1; i <= n; i++) {
      dp[0][i] = dp[1][i] = 1e9;
      for (int j = 0; j < 2; j++) {
        if (dp[j][i - 1] == 1e9) continue;
        for (int k = 0; k < 2; k++) {
          if (dp[j][i - 1] < p[k][i] && p[j][i - 1] < dp[k][i]) {
            dp[k][i] = p[j][i - 1];
            par[k][i] = j;
          }
          if (p[j][i - 1] < p[k][i] && dp[j][i - 1] < dp[k][i]) {
            dp[k][i] = dp[j][i - 1];
            par[k][i] = j;
          }
        }
      }
    }
    if (dp[0][n] == 1e9 && dp[1][n] == 1e9)
      cout << "NO" << '\n';
    else {
      int c = dp[0][n] == 1e9;
      vector<int> ans;
      for (int i = n; i >= 1; i--) {
        ans.push_back(p[c][i]);
        c = par[c][i];
      }
      cout << "YES" << '\n';
      reverse(ans.begin(), ans.end());
      for (int x : ans) cout << x << " ";
      cout << '\n';
    }
  }
}

```