#include <bits/stdc++.h>
using namespace std;
int T, n, p[1000005], dp[1000005][2], g[1000005][2], tans[1000005];
void upd(int &ans, int &ans2, int x, int u, int v, int fl) {
  if (u > v) swap(u, v);
  if (x >= v) {
    if (u < ans) {
      ans = u;
      ans2 = fl;
    }
  } else if (x >= u) {
    if (v < ans) {
      ans = v;
      ans2 = fl;
    }
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
    dp[1][0] = dp[1][1] = -1e9;
    for (int i = 2; i <= n; i++) {
      dp[i][0] = dp[i][1] = 1e9;
      upd(dp[i][0], g[i][0], -p[i], -p[i - 1], dp[i - 1][0], 0);
      upd(dp[i][0], g[i][0], -p[i], p[i - 1], dp[i - 1][1], 1);
      upd(dp[i][1], g[i][1], p[i], -p[i - 1], dp[i - 1][0], 0);
      upd(dp[i][1], g[i][1], p[i], p[i - 1], dp[i - 1][1], 1);
    }
    if (dp[n][1] > 1e8) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    int fl = 1;
    for (int i = n; i > 0; i--) {
      tans[i] = p[i] * (fl * 2 - 1);
      fl = g[i][fl];
    }
    for (int i = 1; i <= n; i++) printf("%d ", tans[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Please formulate 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], dp[1000005][2], g[1000005][2], tans[1000005];
void upd(int &ans, int &ans2, int x, int u, int v, int fl) {
  if (u > v) swap(u, v);
  if (x >= v) {
    if (u < ans) {
      ans = u;
      ans2 = fl;
    }
  } else if (x >= u) {
    if (v < ans) {
      ans = v;
      ans2 = fl;
    }
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
    dp[1][0] = dp[1][1] = -1e9;
    for (int i = 2; i <= n; i++) {
      dp[i][0] = dp[i][1] = 1e9;
      upd(dp[i][0], g[i][0], -p[i], -p[i - 1], dp[i - 1][0], 0);
      upd(dp[i][0], g[i][0], -p[i], p[i - 1], dp[i - 1][1], 1);
      upd(dp[i][1], g[i][1], p[i], -p[i - 1], dp[i - 1][0], 0);
      upd(dp[i][1], g[i][1], p[i], p[i - 1], dp[i - 1][1], 1);
    }
    if (dp[n][1] > 1e8) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    int fl = 1;
    for (int i = n; i > 0; i--) {
      tans[i] = p[i] * (fl * 2 - 1);
      fl = g[i][fl];
    }
    for (int i = 1; i <= n; i++) printf("%d ", tans[i]);
    printf("\n");
  }
  return 0;
}

```