#include <bits/stdc++.h>
using namespace std;
const int N = 1000000;
int t, n;
int p[N];
pair<int, bool> dp[N][2], dp2[N][2];
int ans[N];
void go(int i, bool b) {
  if (i < 0) return;
  ans[i] = b ? -p[i] : p[i];
  go(i - 1, dp[i][b].second);
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> p[0];
    dp[0][0] = dp[0][1] = {-1e9, 0};
    if (n == 1) {
      cout << "YES\n1\n";
      continue;
    }
    for (int i = 1; i < n; i++) {
      cin >> p[i];
      for (int it = 0; it < 2; it++) {
        dp[i][it] =
            min({make_pair(p[i] > p[i - 1] ? dp[i - 1][0].first : n + 1, 0),
                 make_pair(p[i] > -p[i - 1] ? dp[i - 1][1].first : n + 1, 1),
                 make_pair(p[i] > dp[i - 1][0].first ? p[i - 1] : n + 1, 0),
                 make_pair(p[i] > dp[i - 1][1].first ? -p[i - 1] : n + 1, 1)});
        p[i] = -p[i];
      }
    }
    if (dp[n - 1][0].first == dp[n - 1][1].first &&
        dp[n - 1][0].first == n + 1) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    if (dp[n - 1][0].first < 1e9)
      go(n - 1, 0);
    else if (dp[n - 1][1].first < 1e9)
      go(n - 1, 1);
    for (int i = 0; i < n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
}

### Prompt

Your task is to create a Cpp solution 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 = 1000000;
int t, n;
int p[N];
pair<int, bool> dp[N][2], dp2[N][2];
int ans[N];
void go(int i, bool b) {
  if (i < 0) return;
  ans[i] = b ? -p[i] : p[i];
  go(i - 1, dp[i][b].second);
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> p[0];
    dp[0][0] = dp[0][1] = {-1e9, 0};
    if (n == 1) {
      cout << "YES\n1\n";
      continue;
    }
    for (int i = 1; i < n; i++) {
      cin >> p[i];
      for (int it = 0; it < 2; it++) {
        dp[i][it] =
            min({make_pair(p[i] > p[i - 1] ? dp[i - 1][0].first : n + 1, 0),
                 make_pair(p[i] > -p[i - 1] ? dp[i - 1][1].first : n + 1, 1),
                 make_pair(p[i] > dp[i - 1][0].first ? p[i - 1] : n + 1, 0),
                 make_pair(p[i] > dp[i - 1][1].first ? -p[i - 1] : n + 1, 1)});
        p[i] = -p[i];
      }
    }
    if (dp[n - 1][0].first == dp[n - 1][1].first &&
        dp[n - 1][0].first == n + 1) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    if (dp[n - 1][0].first < 1e9)
      go(n - 1, 0);
    else if (dp[n - 1][1].first < 1e9)
      go(n - 1, 1);
    for (int i = 0; i < n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
}

```