description: 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
solution: #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;
  }
  