#include <bits/stdc++.h>
using namespace std;
const int N = 1000001;
map<pair<int, int>, pair<int, int>> parent_state[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> p(n);
    for (auto &i : p) cin >> i;
    fill(parent_state, parent_state + n + 1,
         map<pair<int, int>, pair<int, int>>());
    parent_state[0][{INT_MIN, INT_MIN}];
    for (int i = 0; i < n; ++i) {
      auto it = parent_state[i].begin();
      while (it != parent_state[i].end()) {
        auto cur = it++;
        auto [x, y] = cur->first;
        bool redundant = false;
        for (auto oth = parent_state[i].begin(); oth != cur; ++oth) {
          auto [x_oth, y_oth] = oth->first;
          redundant |= x >= x_oth && y >= y_oth;
        }
        if (redundant) {
          parent_state[i].erase(cur);
          continue;
        }
        for (auto p_nxt : {-p[i], p[i]}) {
          if (p_nxt > x) {
            parent_state[i + 1][{p_nxt, y}] = {x, y};
          } else if (p_nxt > y) {
            parent_state[i + 1][{x, p_nxt}] = {x, y};
          }
        }
      }
    }
    if (parent_state[n].empty()) {
      cout << "NO\n";
      continue;
    }
    auto [x, y] = parent_state[n].begin()->first;
    for (int i = n - 1; i >= 0; --i) {
      auto [x_parent, y_parent] = parent_state[i + 1][{x, y}];
      p[i] = x != x_parent ? x : y;
      x = x_parent, y = y_parent;
    }
    cout << "YES\n";
    for (auto i : p) {
      cout << i << " ";
    }
    cout << "\n";
  }
}

### 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 N = 1000001;
map<pair<int, int>, pair<int, int>> parent_state[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> p(n);
    for (auto &i : p) cin >> i;
    fill(parent_state, parent_state + n + 1,
         map<pair<int, int>, pair<int, int>>());
    parent_state[0][{INT_MIN, INT_MIN}];
    for (int i = 0; i < n; ++i) {
      auto it = parent_state[i].begin();
      while (it != parent_state[i].end()) {
        auto cur = it++;
        auto [x, y] = cur->first;
        bool redundant = false;
        for (auto oth = parent_state[i].begin(); oth != cur; ++oth) {
          auto [x_oth, y_oth] = oth->first;
          redundant |= x >= x_oth && y >= y_oth;
        }
        if (redundant) {
          parent_state[i].erase(cur);
          continue;
        }
        for (auto p_nxt : {-p[i], p[i]}) {
          if (p_nxt > x) {
            parent_state[i + 1][{p_nxt, y}] = {x, y};
          } else if (p_nxt > y) {
            parent_state[i + 1][{x, p_nxt}] = {x, y};
          }
        }
      }
    }
    if (parent_state[n].empty()) {
      cout << "NO\n";
      continue;
    }
    auto [x, y] = parent_state[n].begin()->first;
    for (int i = n - 1; i >= 0; --i) {
      auto [x_parent, y_parent] = parent_state[i + 1][{x, y}];
      p[i] = x != x_parent ? x : y;
      x = x_parent, y = y_parent;
    }
    cout << "YES\n";
    for (auto i : p) {
      cout << i << " ";
    }
    cout << "\n";
  }
}

```