#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int N = 1000 * 1000 + 13;
int n;
int p[N], a[N];
int dp[N][2], pr[N][2];
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < int(n); ++i) scanf("%d", &p[i]);
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(2); ++j) dp[i][j] = INF;
  dp[0][0] = dp[0][1] = -INF;
  for (int i = 0; i < int(n - 1); ++i) {
    bool flag[2];
    flag[0] = flag[1] = false;
    for (int j = 0; j < int(2); ++j) {
      if (dp[i][j] != INF) {
        int x = j ? -p[i] : p[i];
        int y = dp[i][j];
        if (x < y) swap(x, y);
        for (int nj = 0; nj < int(2); ++nj) {
          int z = nj ? -p[i + 1] : p[i + 1];
          if (z > x) {
            if (dp[i + 1][nj] > y) {
              dp[i + 1][nj] = y;
              pr[i + 1][nj] = j;
              flag[nj] = true;
            }
          } else if (z > y && !flag[nj]) {
            if (dp[i + 1][nj] > x) {
              dp[i + 1][nj] = x;
              pr[i + 1][nj] = j;
            }
          }
        }
      }
    }
  }
  int j = -1;
  for (int i = 0; i < int(2); ++i)
    if (dp[n - 1][i] != INF) j = i;
  if (j == -1) {
    puts("NO");
    return;
  }
  for (int i = n - 1; i >= 0; i--) {
    a[i] = j ? -p[i] : p[i];
    j = pr[i][j];
  }
  puts("YES");
  for (int i = 0; i < int(n); ++i) printf("%d ", a[i]);
  puts("");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

### Prompt

Your task is to create 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 INF = 1e9;
const int N = 1000 * 1000 + 13;
int n;
int p[N], a[N];
int dp[N][2], pr[N][2];
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < int(n); ++i) scanf("%d", &p[i]);
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(2); ++j) dp[i][j] = INF;
  dp[0][0] = dp[0][1] = -INF;
  for (int i = 0; i < int(n - 1); ++i) {
    bool flag[2];
    flag[0] = flag[1] = false;
    for (int j = 0; j < int(2); ++j) {
      if (dp[i][j] != INF) {
        int x = j ? -p[i] : p[i];
        int y = dp[i][j];
        if (x < y) swap(x, y);
        for (int nj = 0; nj < int(2); ++nj) {
          int z = nj ? -p[i + 1] : p[i + 1];
          if (z > x) {
            if (dp[i + 1][nj] > y) {
              dp[i + 1][nj] = y;
              pr[i + 1][nj] = j;
              flag[nj] = true;
            }
          } else if (z > y && !flag[nj]) {
            if (dp[i + 1][nj] > x) {
              dp[i + 1][nj] = x;
              pr[i + 1][nj] = j;
            }
          }
        }
      }
    }
  }
  int j = -1;
  for (int i = 0; i < int(2); ++i)
    if (dp[n - 1][i] != INF) j = i;
  if (j == -1) {
    puts("NO");
    return;
  }
  for (int i = n - 1; i >= 0; i--) {
    a[i] = j ? -p[i] : p[i];
    j = pr[i][j];
  }
  puts("YES");
  for (int i = 0; i < int(n); ++i) printf("%d ", a[i]);
  puts("");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

```