#include <bits/stdc++.h>
using namespace std;
using namespace std;
const long long N = 0 ? 15 : 1000005;
int n, a[N];
int dp[N][2][2];
pair<int, int> pre[N][2][2];
int main(int argc, const char *argv[]) {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int(i) = (1); (i) <= (long long)(n); (i)++) scanf("%d", &a[i]);
    dp[0][0][0] = dp[0][0][1] = -1e9;
    dp[0][1][0] = dp[0][1][1] = 1e9;
    a[0] = 1e9;
    for (int(i) = (1); (i) <= (long long)(n); (i)++) {
      for (int(pos) = (0); (pos) <= (long long)(1); (pos)++) {
        for (int(sign) = (0); (sign) <= (long long)(1); (sign)++) {
          int &ans = dp[i][pos][sign];
          ans = 1e9;
          int z = sign ? a[i] : -a[i];
          for (int(psig) = (0); (psig) <= (long long)(1); (psig)++)
            for (int(ppos) = (0); (ppos) <= (long long)(1); (ppos)++) {
              int x = psig ? a[i - 1] : -a[i - 1];
              int y = dp[i - 1][ppos][psig];
              if (ppos) {
                swap(x, y);
              }
              if (z > x && !pos) {
                if (ans > y) {
                  ans = y;
                  pre[i][pos][sign] = {ppos, psig};
                }
              } else if (z > y && pos) {
                if (ans > x) {
                  ans = x;
                  pre[i][pos][sign] = {ppos, psig};
                }
              }
            }
        }
      }
    }
    int pos = -1, sign = -1;
    for (int(p) = (0); (p) <= (long long)(1); (p)++)
      for (int(second) = (0); (second) <= (long long)(1); (second)++)
        if (dp[n][p][second] != 1e9) {
          pos = p;
          sign = second;
        }
    if (pos == -1) {
      printf("NO\n");
      continue;
    }
    for (long long(i) = (n); (i) >= (1); --(i)) {
      a[i] = sign ? a[i] : -a[i];
      pair<int, int> nxt = pre[i][pos][sign];
      pos = nxt.first;
      sign = nxt.second;
    }
    printf("YES\n");
    for (int(i) = (1); (i) <= (long long)(n); (i)++) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
using namespace std;
const long long N = 0 ? 15 : 1000005;
int n, a[N];
int dp[N][2][2];
pair<int, int> pre[N][2][2];
int main(int argc, const char *argv[]) {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int(i) = (1); (i) <= (long long)(n); (i)++) scanf("%d", &a[i]);
    dp[0][0][0] = dp[0][0][1] = -1e9;
    dp[0][1][0] = dp[0][1][1] = 1e9;
    a[0] = 1e9;
    for (int(i) = (1); (i) <= (long long)(n); (i)++) {
      for (int(pos) = (0); (pos) <= (long long)(1); (pos)++) {
        for (int(sign) = (0); (sign) <= (long long)(1); (sign)++) {
          int &ans = dp[i][pos][sign];
          ans = 1e9;
          int z = sign ? a[i] : -a[i];
          for (int(psig) = (0); (psig) <= (long long)(1); (psig)++)
            for (int(ppos) = (0); (ppos) <= (long long)(1); (ppos)++) {
              int x = psig ? a[i - 1] : -a[i - 1];
              int y = dp[i - 1][ppos][psig];
              if (ppos) {
                swap(x, y);
              }
              if (z > x && !pos) {
                if (ans > y) {
                  ans = y;
                  pre[i][pos][sign] = {ppos, psig};
                }
              } else if (z > y && pos) {
                if (ans > x) {
                  ans = x;
                  pre[i][pos][sign] = {ppos, psig};
                }
              }
            }
        }
      }
    }
    int pos = -1, sign = -1;
    for (int(p) = (0); (p) <= (long long)(1); (p)++)
      for (int(second) = (0); (second) <= (long long)(1); (second)++)
        if (dp[n][p][second] != 1e9) {
          pos = p;
          sign = second;
        }
    if (pos == -1) {
      printf("NO\n");
      continue;
    }
    for (long long(i) = (n); (i) >= (1); --(i)) {
      a[i] = sign ? a[i] : -a[i];
      pair<int, int> nxt = pre[i][pos][sign];
      pos = nxt.first;
      sign = nxt.second;
    }
    printf("YES\n");
    for (int(i) = (1); (i) <= (long long)(n); (i)++) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}

```