#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m, t, it, a[1005000], fa[10005000];
bool op[10050000];
struct SB {
  int id, x, y;
  bool operator<(const SB a) const { return (x == a.x) ? y < a.y : x < a.x; }
};
vector<SB> v1, v2;
void add(int x, int y, int z, int b) {
  fa[++it] = x;
  op[it] = b;
  v2.push_back({it, y, z});
}
int main() {
  ios::sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> n;
    it = 0;
    for (i = 1; i <= n; i++) cin >> a[i];
    v1.push_back({0, -11451419, -11451419});
    for (i = 1; i <= n; i++) {
      v2.clear();
      for (auto &[id, x, y] : v1) {
        if (x < -a[i])
          add(id, -a[i], y, 1);
        else if (y < -a[i])
          add(id, x, -a[i], 1);
        if (x < a[i])
          add(id, a[i], y, 0);
        else if (y < a[i])
          add(id, x, a[i], 0);
      }
      if (v2.empty()) {
        cout << "NO\n";
        goto ccc;
      }
      v1.clear();
      k = 1e9;
      sort(v2.begin(), v2.end());
      for (auto &i : v2) {
        if (i.y < k) {
          k = i.y;
          v1.push_back(i);
        }
      }
    }
    i = (*v1.begin()).id;
    j = n;
    while (i) {
      if (op[i]) a[j] *= -1;
      j--;
      i = fa[i];
    }
    cout << "YES\n";
    for (i = 1; i <= n; i++) cout << a[i] << ' ';
    cout << '\n';
  ccc:;
  }
}

### Prompt

In CPP, your task is to solve 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 i, j, k, n, m, t, it, a[1005000], fa[10005000];
bool op[10050000];
struct SB {
  int id, x, y;
  bool operator<(const SB a) const { return (x == a.x) ? y < a.y : x < a.x; }
};
vector<SB> v1, v2;
void add(int x, int y, int z, int b) {
  fa[++it] = x;
  op[it] = b;
  v2.push_back({it, y, z});
}
int main() {
  ios::sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> n;
    it = 0;
    for (i = 1; i <= n; i++) cin >> a[i];
    v1.push_back({0, -11451419, -11451419});
    for (i = 1; i <= n; i++) {
      v2.clear();
      for (auto &[id, x, y] : v1) {
        if (x < -a[i])
          add(id, -a[i], y, 1);
        else if (y < -a[i])
          add(id, x, -a[i], 1);
        if (x < a[i])
          add(id, a[i], y, 0);
        else if (y < a[i])
          add(id, x, a[i], 0);
      }
      if (v2.empty()) {
        cout << "NO\n";
        goto ccc;
      }
      v1.clear();
      k = 1e9;
      sort(v2.begin(), v2.end());
      for (auto &i : v2) {
        if (i.y < k) {
          k = i.y;
          v1.push_back(i);
        }
      }
    }
    i = (*v1.begin()).id;
    j = n;
    while (i) {
      if (op[i]) a[j] *= -1;
      j--;
      i = fa[i];
    }
    cout << "YES\n";
    for (i = 1; i <= n; i++) cout << a[i] << ' ';
    cout << '\n';
  ccc:;
  }
}

```