#include <bits/stdc++.h>
using namespace std;
int a[1000000 + 5];
int f[1000000 + 5][2][2];
pair<int, int> g[1000000 + 5][2][2];
void print(int n, int i, int j) {
  if (n == 0) return;
  print(n - 1, g[n][i][j].first, g[n][i][j].second);
  if (j == 0)
    printf("%d ", a[n]);
  else
    printf("%d ", -a[n]);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, i, j, k, l;
    scanf("%d", &n);
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (i = 0; i <= n + 1; i++)
      for (j = 0; j < 2; j++)
        for (k = 0; k < 2; k++) f[i][j][k] = 1e9;
    f[1][0][0] = f[1][0][1] = -1e9;
    for (i = 1; i < n; i++)
      for (j = 0; j < 2; j++)
        for (k = 0; k < 2; k++) {
          int x = a[i], y = f[i][j][k];
          if (k) x = -x;
          if (j) swap(x, y);
          for (l = 0; l < 2; l++) {
            int z = a[i + 1];
            if (l) z = -z;
            if (z < y) continue;
            if (z < x) {
              if (f[i + 1][1][l] > x)
                f[i + 1][1][l] = x, g[i + 1][1][l] = make_pair(j, k);
            } else {
              if (f[i + 1][0][l] > y)
                f[i + 1][0][l] = y, g[i + 1][0][l] = make_pair(j, k);
            }
          }
        }
    bool flag = false;
    for (i = 0; i < 2 && !flag; i++)
      for (j = 0; j < 2 && !flag; j++)
        if (f[n][i][j] <= n && !flag) {
          printf("YES\n");
          print(n, i, j);
          flag = true;
          break;
        }
    if (!flag) printf("NO");
    printf("\n");
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 a[1000000 + 5];
int f[1000000 + 5][2][2];
pair<int, int> g[1000000 + 5][2][2];
void print(int n, int i, int j) {
  if (n == 0) return;
  print(n - 1, g[n][i][j].first, g[n][i][j].second);
  if (j == 0)
    printf("%d ", a[n]);
  else
    printf("%d ", -a[n]);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, i, j, k, l;
    scanf("%d", &n);
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (i = 0; i <= n + 1; i++)
      for (j = 0; j < 2; j++)
        for (k = 0; k < 2; k++) f[i][j][k] = 1e9;
    f[1][0][0] = f[1][0][1] = -1e9;
    for (i = 1; i < n; i++)
      for (j = 0; j < 2; j++)
        for (k = 0; k < 2; k++) {
          int x = a[i], y = f[i][j][k];
          if (k) x = -x;
          if (j) swap(x, y);
          for (l = 0; l < 2; l++) {
            int z = a[i + 1];
            if (l) z = -z;
            if (z < y) continue;
            if (z < x) {
              if (f[i + 1][1][l] > x)
                f[i + 1][1][l] = x, g[i + 1][1][l] = make_pair(j, k);
            } else {
              if (f[i + 1][0][l] > y)
                f[i + 1][0][l] = y, g[i + 1][0][l] = make_pair(j, k);
            }
          }
        }
    bool flag = false;
    for (i = 0; i < 2 && !flag; i++)
      for (j = 0; j < 2 && !flag; j++)
        if (f[n][i][j] <= n && !flag) {
          printf("YES\n");
          print(n, i, j);
          flag = true;
          break;
        }
    if (!flag) printf("NO");
    printf("\n");
  }
}

```