#include <bits/stdc++.h>
using namespace std;
int T, n, p[1000005], q[1000005];
int f[1000005][2][2], g[1000005][2][2];
void sol() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++)
    f[i][0][0] = f[i][0][1] = f[i][1][0] = f[i][1][1] = n + 1;
  f[1][0][0] = f[1][1][0] = -n - 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++)
        if (f[i][j][k] <= n) {
          int x = j ? p[i] : -p[i], y = f[i][j][k];
          if (k) swap(x, y);
          for (int l = 0; l < 2; l++) {
            int z = l ? p[i + 1] : -p[i + 1];
            if (z < y) continue;
            if (z < x) {
              if (f[i + 1][l][1] > x) {
                f[i + 1][l][1] = x;
                g[i + 1][l][1] = j << 1 | k;
              }
            } else {
              if (f[i + 1][l][0] > y) {
                f[i + 1][l][0] = y;
                g[i + 1][l][0] = j << 1 | k;
              }
            }
          }
        }
  int t1 = -1, t2 = -1;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      if (f[n][i][j] <= n) t1 = i, t2 = j;
  if (t1 < 0) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = n; i; i--) {
    q[i] = t1 ? p[i] : -p[i];
    if (i == 1) break;
    int msk = g[i][t1][t2];
    t1 = msk >> 1, t2 = msk & 1;
  }
  for (int i = 1; i <= n; i++) printf("%d ", q[i]);
  puts("");
}
int main() {
  scanf("%d", &T);
  while (T--) sol();
}

### 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;
int T, n, p[1000005], q[1000005];
int f[1000005][2][2], g[1000005][2][2];
void sol() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++)
    f[i][0][0] = f[i][0][1] = f[i][1][0] = f[i][1][1] = n + 1;
  f[1][0][0] = f[1][1][0] = -n - 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++)
        if (f[i][j][k] <= n) {
          int x = j ? p[i] : -p[i], y = f[i][j][k];
          if (k) swap(x, y);
          for (int l = 0; l < 2; l++) {
            int z = l ? p[i + 1] : -p[i + 1];
            if (z < y) continue;
            if (z < x) {
              if (f[i + 1][l][1] > x) {
                f[i + 1][l][1] = x;
                g[i + 1][l][1] = j << 1 | k;
              }
            } else {
              if (f[i + 1][l][0] > y) {
                f[i + 1][l][0] = y;
                g[i + 1][l][0] = j << 1 | k;
              }
            }
          }
        }
  int t1 = -1, t2 = -1;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      if (f[n][i][j] <= n) t1 = i, t2 = j;
  if (t1 < 0) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = n; i; i--) {
    q[i] = t1 ? p[i] : -p[i];
    if (i == 1) break;
    int msk = g[i][t1][t2];
    t1 = msk >> 1, t2 = msk & 1;
  }
  for (int i = 1; i <= n; i++) printf("%d ", q[i]);
  puts("");
}
int main() {
  scanf("%d", &T);
  while (T--) sol();
}

```