#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n + 1, 0);
    for (int i = 1; i <= n; i++) cin >> a[i];
    const int inf = (1 << 30);
    vector<array<int, 2>> dp(n + 1, {inf, inf});
    vector<array<int, 2>> p(n + 1, {-1, -1});
    dp[1] = {-inf, -inf};
    auto upd = [&](int i, int j, int lim, int u) {
      if (lim < dp[i][j]) {
        dp[i][j] = lim;
        p[i][j] = u;
      }
    };
    for (int i = 1; i < n; i++) {
      for (int j = 0; j < 2; j++) {
        if (dp[i][j] == inf) continue;
        int x = (j ? -a[i] : a[i]);
        for (int k = 0; k < 2; k++) {
          int y = (k ? -a[i + 1] : a[i + 1]);
          if (x < y) upd(i + 1, k, dp[i][j], j);
          if (dp[i][j] < y) upd(i + 1, k, x, j);
        }
      }
    }
    if (dp[n][0] != inf) {
      int s = 0;
      cout << "YES\n";
      vector<int> res(n + 1, 0);
      for (int i = n; i >= 1; i--) {
        res[i] = (s ? -a[i] : a[i]);
        s = p[i][s];
      }
      for (int i = 1; i <= n; i++) cout << res[i] << " \n"[i == n];
    } else
      cout << "NO\n";
  }
  return 0;
}

### Prompt

Generate 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n + 1, 0);
    for (int i = 1; i <= n; i++) cin >> a[i];
    const int inf = (1 << 30);
    vector<array<int, 2>> dp(n + 1, {inf, inf});
    vector<array<int, 2>> p(n + 1, {-1, -1});
    dp[1] = {-inf, -inf};
    auto upd = [&](int i, int j, int lim, int u) {
      if (lim < dp[i][j]) {
        dp[i][j] = lim;
        p[i][j] = u;
      }
    };
    for (int i = 1; i < n; i++) {
      for (int j = 0; j < 2; j++) {
        if (dp[i][j] == inf) continue;
        int x = (j ? -a[i] : a[i]);
        for (int k = 0; k < 2; k++) {
          int y = (k ? -a[i + 1] : a[i + 1]);
          if (x < y) upd(i + 1, k, dp[i][j], j);
          if (dp[i][j] < y) upd(i + 1, k, x, j);
        }
      }
    }
    if (dp[n][0] != inf) {
      int s = 0;
      cout << "YES\n";
      vector<int> res(n + 1, 0);
      for (int i = n; i >= 1; i--) {
        res[i] = (s ? -a[i] : a[i]);
        s = p[i][s];
      }
      for (int i = 1; i <= n; i++) cout << res[i] << " \n"[i == n];
    } else
      cout << "NO\n";
  }
  return 0;
}

```