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(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> v(n);
      for (int &x : v) cin >> x;
      vector<vector<int>> dp(n, vector<int>(2, (int)1e9)),
          elozo(n, vector<int>(2, 0));
      dp[0] = {-(int)1e9, -(int)1e9};
      for (int i = 1; i < n; i++) {
        dp[i][0] = dp[i - 1][1];
        if (v[i] > dp[i - 1][1]) dp[i][0] = min(dp[i][0], -v[i - 1]);
        elozo[i][0] = 1;
        if (v[i] > dp[i - 1][0] && dp[i][0] > v[i - 1]) {
          dp[i][0] = v[i - 1];
          elozo[i][0] = 0;
        }
        if (v[i] > v[i - 1] && dp[i][0] > dp[i - 1][0]) {
          dp[i][0] = dp[i - 1][0];
          elozo[i][0] = 0;
        }
        if (-v[i] > dp[i - 1][0] && dp[i][1] > v[i - 1]) {
          dp[i][1] = v[i - 1];
          elozo[i][1] = 0;
        }
        if (-v[i] > dp[i - 1][1] && dp[i][1] > -v[i - 1]) {
          dp[i][1] = -v[i - 1];
          elozo[i][1] = 1;
        }
        if (-v[i] > -v[i - 1] && dp[i][1] > dp[i - 1][1]) {
          dp[i][1] = dp[i - 1][1];
          elozo[i][1] = 1;
        }
      }
      if (dp[n - 1][0] == (int)1e9) {
        cout << "NO\n";
        continue;
      }
      cout << "YES\n";
      int x = 0;
      for (int i = n - 1; i >= 0; i--) {
        if (x) v[i] = -v[i];
        x = elozo[i][x];
      }
      for (int &x : v) cout << x << ' ';
      cout << '\n';
    }
    return 0;
  }
  