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 N = 1000000;
  int t, n;
  int p[N];
  pair<int, bool> dp[N][2], dp2[N][2];
  int ans[N];
  void go(int i, bool b) {
    if (i < 0) return;
    ans[i] = b ? -p[i] : p[i];
    go(i - 1, dp[i][b].second);
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> t;
    while (t--) {
      cin >> n;
      cin >> p[0];
      dp[0][0] = dp[0][1] = {-1e9, 0};
      if (n == 1) {
        cout << "YES\n1\n";
        continue;
      }
      for (int i = 1; i < n; i++) {
        cin >> p[i];
        for (int it = 0; it < 2; it++) {
          dp[i][it] =
              min({make_pair(p[i] > p[i - 1] ? dp[i - 1][0].first : n + 1, 0),
                   make_pair(p[i] > -p[i - 1] ? dp[i - 1][1].first : n + 1, 1),
                   make_pair(p[i] > dp[i - 1][0].first ? p[i - 1] : n + 1, 0),
                   make_pair(p[i] > dp[i - 1][1].first ? -p[i - 1] : n + 1, 1)});
          p[i] = -p[i];
        }
      }
      if (dp[n - 1][0].first == dp[n - 1][1].first &&
          dp[n - 1][0].first == n + 1) {
        cout << "NO\n";
        continue;
      }
      cout << "YES\n";
      if (dp[n - 1][0].first < 1e9)
        go(n - 1, 0);
      else if (dp[n - 1][1].first < 1e9)
        go(n - 1, 1);
      for (int i = 0; i < n; i++) cout << ans[i] << " ";
      cout << "\n";
    }
  }
  