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 = 2e5 + 100;
  const int mod = 998244353;
  signed main() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> p(n);
      for (int i = 0; i < n; i++) {
        cin >> p[i];
      }
      int dp[n][2][2];
      vector<int> pred[n][2][2];
      for (int i = 0; i < n; i++) {
        for (int first = 0; first < 2; first++) {
          for (int second = 0; second < 2; second++) {
            dp[i][first][second] = 1e9;
            pred[i][first][second] = {-1, -1, -1, -1};
          }
        }
      }
      dp[0][0][0] = -1e9;
      dp[0][0][1] = -1e9;
      pred[0][0][0] = {-1, -1, -1, -p[0]};
      pred[0][0][1] = {-1, -1, -1, p[0]};
      for (int i = 1; i < n; i++) {
        for (int first = 0; first < 2; first++) {
          int a, b;
          for (int second = 0; second < 2; second++) {
            if (dp[i - 1][first][second] == 1e9) {
              continue;
            }
            if (first == 0) {
              b = dp[i - 1][first][second];
              a = p[i - 1];
              if (second == 0) {
                a *= -1;
              }
            } else {
              b = p[i - 1];
              a = dp[i - 1][first][second];
              if (second == 0) {
                b *= -1;
              }
            }
            if (p[i] > a) {
              if (dp[i][0][1] > b) {
                dp[i][0][1] = b;
                pred[i][0][1] = {i - 1, first, second, p[i]};
              }
            } else if (p[i] > b) {
              if (dp[i][1][1] > a) {
                dp[i][1][1] = a;
                pred[i][1][1] = {i - 1, first, second, p[i]};
              }
            }
            if (-p[i] > a) {
              if (dp[i][0][0] > b) {
                dp[i][0][0] = b;
                pred[i][0][0] = {i - 1, first, second, -p[i]};
              }
            } else if (-p[i] > b) {
              if (dp[i][1][0] > a) {
                dp[i][1][0] = a;
                pred[i][1][0] = {i - 1, first, second, -p[i]};
              }
            }
          }
        }
      }
      int fl = 0;
      vector<int> ans(n);
      for (int first = 0; first < 2; first++) {
        for (int second = 0; second < 2; second++) {
          if (dp[n - 1][first][second] < 1e9) {
            int i, x1, y1;
            i = n - 1;
            x1 = first;
            y1 = second;
            while (i >= 0) {
              int x2, y2;
              x2 = pred[i][x1][y1][1];
              y2 = pred[i][x1][y1][2];
              ans[i] = pred[i][x1][y1][3];
              x1 = x2;
              y1 = y2;
              i--;
            }
            fl = 1;
          }
        }
      }
      if (!fl) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        for (int i = 0; i < n; i++) {
          cout << ans[i] << " ";
        }
        cout << "\n";
      }
    }
    return 0;
  }
  