#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
const int INf = 0x7f7f7f7f;
int n, a[N], f[N][2], g[N][2];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
      f[i][0] = f[i][1] = INf;
    }
    f[1][0] = f[1][1] = -INf;
    for (int i = 2; i <= n; i++) {
      if (a[i] > a[i - 1] && f[i][0] > f[i - 1][0]) {
        f[i][0] = f[i - 1][0];
        g[i][0] = 0;
      }
      if (a[i] > f[i - 1][0] && f[i][0] > a[i - 1]) {
        f[i][0] = a[i - 1];
        g[i][0] = 0;
      }
      if (a[i] > -a[i - 1] && f[i][0] > f[i - 1][1]) {
        f[i][0] = f[i - 1][1];
        g[i][0] = 1;
      }
      if (a[i] > f[i - 1][1] && f[i][0] > -a[i - 1]) {
        f[i][0] = -a[i - 1];
        g[i][0] = 1;
      }
      if (-a[i] > a[i - 1] && f[i][1] > f[i - 1][0]) {
        f[i][1] = f[i - 1][0];
        g[i][1] = 0;
      }
      if (-a[i] > f[i - 1][0] && f[i][1] > a[i - 1]) {
        f[i][1] = a[i - 1];
        g[i][1] = 0;
      }
      if (-a[i] > -a[i - 1] && f[i][1] > f[i - 1][1]) {
        f[i][1] = f[i - 1][1];
        g[i][1] = 1;
      }
      if (-a[i] > f[i - 1][1] && f[i][1] > -a[i - 1]) {
        f[i][1] = -a[i - 1];
        g[i][1] = 1;
      }
    }
    if (f[n][0] == INf && f[n][1] == INf) {
      printf("NO\n");
    } else {
      printf("YES\n");
      int u = f[n][0] == INf ? 1 : 0;
      for (int i = n; i >= 1; i--) {
        a[i] = u ? -a[i] : a[i];
        u = g[i][u];
      }
      for (int i = 1; i <= n; i++) {
        printf("%d ", a[i]);
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 = 1000010;
const int INf = 0x7f7f7f7f;
int n, a[N], f[N][2], g[N][2];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
      f[i][0] = f[i][1] = INf;
    }
    f[1][0] = f[1][1] = -INf;
    for (int i = 2; i <= n; i++) {
      if (a[i] > a[i - 1] && f[i][0] > f[i - 1][0]) {
        f[i][0] = f[i - 1][0];
        g[i][0] = 0;
      }
      if (a[i] > f[i - 1][0] && f[i][0] > a[i - 1]) {
        f[i][0] = a[i - 1];
        g[i][0] = 0;
      }
      if (a[i] > -a[i - 1] && f[i][0] > f[i - 1][1]) {
        f[i][0] = f[i - 1][1];
        g[i][0] = 1;
      }
      if (a[i] > f[i - 1][1] && f[i][0] > -a[i - 1]) {
        f[i][0] = -a[i - 1];
        g[i][0] = 1;
      }
      if (-a[i] > a[i - 1] && f[i][1] > f[i - 1][0]) {
        f[i][1] = f[i - 1][0];
        g[i][1] = 0;
      }
      if (-a[i] > f[i - 1][0] && f[i][1] > a[i - 1]) {
        f[i][1] = a[i - 1];
        g[i][1] = 0;
      }
      if (-a[i] > -a[i - 1] && f[i][1] > f[i - 1][1]) {
        f[i][1] = f[i - 1][1];
        g[i][1] = 1;
      }
      if (-a[i] > f[i - 1][1] && f[i][1] > -a[i - 1]) {
        f[i][1] = -a[i - 1];
        g[i][1] = 1;
      }
    }
    if (f[n][0] == INf && f[n][1] == INf) {
      printf("NO\n");
    } else {
      printf("YES\n");
      int u = f[n][0] == INf ? 1 : 0;
      for (int i = n; i >= 1; i--) {
        a[i] = u ? -a[i] : a[i];
        u = g[i][u];
      }
      for (int i = 1; i <= n; i++) {
        printf("%d ", a[i]);
      }
      printf("\n");
    }
  }
  return 0;
}

```