#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5;
int n, A[N], P[N], f[N];
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &A[i]);
  for (int i = 1; i < 2 * n; ++i) P[i] = -1, f[i] = n + 1;
  f[0] = f[1] = -n;
  for (int i = 1; i < n; ++i)
    for (int x = 0; x < 2; ++x)
      for (int y = 0; y < 2; ++y) {
        int v[2] = {f[i * 2 - 2 + y], A[i - 1] * (2 * y - 1)};
        for (int j = 0; j < 2; ++j)
          if (v[j] <= A[i] * (2 * x - 1) && v[(j == 0)] < f[i * 2 + x])
            f[i * 2 + x] = v[!j], P[i * 2 + x] = i * 2 - 2 + y;
      }
  if (f[2 * n - 1] <= n) {
    puts("YES");
    for (int i = n - 1, x = 2 * n - 1; ~i; --i, x = P[x])
      if (x % 2 == 0) A[i] *= -1;
    for (int i = 0; i < n; ++i) printf("%d ", A[i]);
    puts("");
  } else {
    puts("NO");
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  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;
const int N = 2e6 + 5;
int n, A[N], P[N], f[N];
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &A[i]);
  for (int i = 1; i < 2 * n; ++i) P[i] = -1, f[i] = n + 1;
  f[0] = f[1] = -n;
  for (int i = 1; i < n; ++i)
    for (int x = 0; x < 2; ++x)
      for (int y = 0; y < 2; ++y) {
        int v[2] = {f[i * 2 - 2 + y], A[i - 1] * (2 * y - 1)};
        for (int j = 0; j < 2; ++j)
          if (v[j] <= A[i] * (2 * x - 1) && v[(j == 0)] < f[i * 2 + x])
            f[i * 2 + x] = v[!j], P[i * 2 + x] = i * 2 - 2 + y;
      }
  if (f[2 * n - 1] <= n) {
    puts("YES");
    for (int i = n - 1, x = 2 * n - 1; ~i; --i, x = P[x])
      if (x % 2 == 0) A[i] *= -1;
    for (int i = 0; i < n; ++i) printf("%d ", A[i]);
    puts("");
  } else {
    puts("NO");
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```