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;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<vector<int>> a(n, vector<int>(2));
      vector<vector<pair<int, int>>> dp(n, vector<pair<int, int>>(2));
      for (int i = 0; i < n; i++) {
        cin >> a[i][1];
        a[i][0] = -a[i][1];
        if (i == 0) {
          dp[i][0] = dp[i][1] = {-n - 1, 0};
        } else {
          for (int j = 0; j < 2; j++) {
            dp[i][j] = {n + 1, 0};
            for (int k = 0; k < 2; k++) {
              if (a[i - 1][k] < a[i][j]) {
                dp[i][j] = min(dp[i][j], {dp[i - 1][k].first, k});
              }
              if (dp[i - 1][k].first < a[i][j]) {
                dp[i][j] = min(dp[i][j], {a[i - 1][k], k});
              }
            }
          }
        }
      }
      for (int cur = 0; cur < 2; cur++) {
        if (dp[n - 1][cur].first <= n) {
          cout << "YES\n";
          for (int i = n - 1; i >= 0; i--) {
            a[i][0] = a[i][cur];
            cur = dp[i][cur].second;
          }
          for (int i = 0; i < n; i++) {
            cout << a[i][0] << " \n"[i == n - 1];
          }
          goto hell;
        }
      }
      cout << "NO\n";
    hell:
      continue;
    }
  }
  