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 inf = 1e9;
  int main() {
    std::ios_base::sync_with_stdio(false);
    int T;
    cin >> T;
    while (T--) {
      int n;
      cin >> n;
      vector<int> a(n + 1);
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
      }
      vector<vector<int>> f(n + 1, vector<int>(4, inf));
      vector<vector<int>> pre(n + 1, vector<int>(4, -1));
      f[1][0] = f[1][1] = -inf;
      for (int i = 2; i <= n; i++) {
        f[i] = vector<int>(4, inf);
        vector<tuple<int, int, int>> b;
        for (int j = 0; j < 4; j++) {
          int x = (j % 2 ? -1 : 1) * a[i - 1];
          if (f[i - 1][j] != inf) {
            int y = f[i - 1][j];
            if (x < y) {
              swap(x, y);
            }
            b.emplace_back(j, x, y);
          }
        }
        for (int j = 0; j < 4; j++) {
          int u = (j % 2 ? -1 : 1) * a[i];
          for (auto [p, x, y] : b) {
            if (j < 2 && u > x) {
              if (y < f[i][j]) {
                f[i][j] = y;
                pre[i][j] = p;
              }
            }
            if (j >= 2 && u > y && u < x) {
              if (x < f[i][j]) {
                f[i][j] = x;
                pre[i][j] = p;
              }
            }
          }
        }
      }
      bool ok = false;
      vector<int> ans(n + 1);
      for (int j = 0; j < 4; j++) {
        if (f[n][j] < inf) {
          int p = j;
          ok = true;
          for (int i = n; i >= 1; i--) {
            ans[i] = (p % 2 ? -1 : 1) * a[i];
            p = pre[i][p];
          }
          break;
        }
      }
      cout << (ok ? "YES" : "NO") << endl;
      if (ok) {
        for (int i = 1; i <= n; i++) {
          cout << ans[i] << " ";
        }
        cout << endl;
      }
    }
    return 0;
  }
  