#include <bits/stdc++.h>
using namespace std;
const int c = 1000005;
int w, n, t[c];
vector<pair<pair<int, int>, pair<int, int> > > sz[c];
void pb(int a, int b, int c, int d, int e) {
  sz[a].push_back({{b, c}, {d, e}});
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> w;
  while (w--) {
    cin >> n;
    pb(0, -c, -c, 0, 0);
    for (int i = 1; i <= n; i++) {
      cin >> t[i];
      for (auto p : sz[i - 1]) {
        int x = p.first.first, y = p.first.second;
        if (t[i] > x) {
          pb(n + 1, t[i], y, x, y);
        } else if (t[i] > y) {
          pb(n + 1, x, t[i], x, y);
        }
        if (-t[i] > x) {
          pb(n + 1, -t[i], y, x, y);
        } else if (-t[i] > y) {
          pb(n + 1, x, -t[i], x, y);
        }
      }
      sort(sz[n + 1].begin(), sz[n + 1].end());
      for (auto x : sz[n + 1]) {
        if (sz[i].size() == 0 || x.first.second < sz[i].back().first.second) {
          sz[i].push_back(x);
        }
      }
      sz[n + 1].clear();
    }
    if (sz[n].size() > 0) {
      cout << "YES\n";
      vector<int> res;
      pair<int, int> p = sz[n][0].first;
      for (int i = n; i >= 1; i--) {
        for (auto j : sz[i]) {
          if (j.first == p) {
            res.push_back(abs(p.first) == t[i] ? p.first : p.second);
            p = j.second;
            break;
          }
        }
      }
      reverse(res.begin(), res.end());
      for (auto x : res) {
        cout << x << " ";
      }
      cout << "\n";
    } else {
      cout << "NO\n";
    }
    for (int i = 0; i <= n; i++) {
      sz[i].clear();
    }
  }
  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 c = 1000005;
int w, n, t[c];
vector<pair<pair<int, int>, pair<int, int> > > sz[c];
void pb(int a, int b, int c, int d, int e) {
  sz[a].push_back({{b, c}, {d, e}});
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> w;
  while (w--) {
    cin >> n;
    pb(0, -c, -c, 0, 0);
    for (int i = 1; i <= n; i++) {
      cin >> t[i];
      for (auto p : sz[i - 1]) {
        int x = p.first.first, y = p.first.second;
        if (t[i] > x) {
          pb(n + 1, t[i], y, x, y);
        } else if (t[i] > y) {
          pb(n + 1, x, t[i], x, y);
        }
        if (-t[i] > x) {
          pb(n + 1, -t[i], y, x, y);
        } else if (-t[i] > y) {
          pb(n + 1, x, -t[i], x, y);
        }
      }
      sort(sz[n + 1].begin(), sz[n + 1].end());
      for (auto x : sz[n + 1]) {
        if (sz[i].size() == 0 || x.first.second < sz[i].back().first.second) {
          sz[i].push_back(x);
        }
      }
      sz[n + 1].clear();
    }
    if (sz[n].size() > 0) {
      cout << "YES\n";
      vector<int> res;
      pair<int, int> p = sz[n][0].first;
      for (int i = n; i >= 1; i--) {
        for (auto j : sz[i]) {
          if (j.first == p) {
            res.push_back(abs(p.first) == t[i] ? p.first : p.second);
            p = j.second;
            break;
          }
        }
      }
      reverse(res.begin(), res.end());
      for (auto x : res) {
        cout << x << " ";
      }
      cout << "\n";
    } else {
      cout << "NO\n";
    }
    for (int i = 0; i <= n; i++) {
      sz[i].clear();
    }
  }
  return 0;
}

```