#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, p[N][2], dp[N][2][2];
pair<int, int> g[N][2][2];
void cmin(int& x, int y) {
  if (x > y) x = y;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i][1]), p[i][0] = -p[i][1];
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 1; j++)
      for (int k = 0; k <= 1; k++) dp[i][j][k] = n + 1;
  dp[1][0][0] = dp[1][0][1] = -n - 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= 1; j++)
      for (int k = 0; k <= 1; k++)
        for (int l = 0; l <= 1; l++) {
          int x = (j == 0) ? p[i][k] : dp[i][j][k],
              y = (j == 1) ? p[i][k] : dp[i][j][k];
          if (x > p[i + 1][l] && p[i + 1][l] > y && dp[i + 1][1][l] > x)
            dp[i + 1][1][l] = x, g[i + 1][1][l] = {j, k};
          if (p[i + 1][l] > x && dp[i + 1][0][l] > y)
            dp[i + 1][0][l] = y, g[i + 1][0][l] = {j, k};
        }
  vector<int> ans;
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j <= 1; j++)
      if (dp[n][i][j] != n + 1) {
        puts("YES");
        pair<int, int> now = g[n][i][j];
        int tmp = n - 1;
        ans.push_back(p[n][j]);
        while (tmp) {
          ans.push_back(p[tmp][now.second]);
          now = g[tmp][now.first][now.second];
          tmp--;
        }
        for (int i = ans.size() - 1; i >= 0; i--) printf("%d ", ans[i]);
        puts("");
        return;
      }
  puts("NO");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 1e6 + 5;
int n, p[N][2], dp[N][2][2];
pair<int, int> g[N][2][2];
void cmin(int& x, int y) {
  if (x > y) x = y;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i][1]), p[i][0] = -p[i][1];
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 1; j++)
      for (int k = 0; k <= 1; k++) dp[i][j][k] = n + 1;
  dp[1][0][0] = dp[1][0][1] = -n - 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= 1; j++)
      for (int k = 0; k <= 1; k++)
        for (int l = 0; l <= 1; l++) {
          int x = (j == 0) ? p[i][k] : dp[i][j][k],
              y = (j == 1) ? p[i][k] : dp[i][j][k];
          if (x > p[i + 1][l] && p[i + 1][l] > y && dp[i + 1][1][l] > x)
            dp[i + 1][1][l] = x, g[i + 1][1][l] = {j, k};
          if (p[i + 1][l] > x && dp[i + 1][0][l] > y)
            dp[i + 1][0][l] = y, g[i + 1][0][l] = {j, k};
        }
  vector<int> ans;
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j <= 1; j++)
      if (dp[n][i][j] != n + 1) {
        puts("YES");
        pair<int, int> now = g[n][i][j];
        int tmp = n - 1;
        ans.push_back(p[n][j]);
        while (tmp) {
          ans.push_back(p[tmp][now.second]);
          now = g[tmp][now.first][now.second];
          tmp--;
        }
        for (int i = ans.size() - 1; i >= 0; i--) printf("%d ", ans[i]);
        puts("");
        return;
      }
  puts("NO");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
}

```