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(int argc, char const *argv[]) {
    int t, n;
    cin >> t;
    vector<int> a;
    vector<int> f[2][2];
    vector<pair<int, int>> trace[2][2];
    while (t--) {
      cin >> n;
      a.resize(n);
      for (auto &u : a) cin >> u;
      for (int i = 0; i < 2; ++i)
        for (int j = 0; j < 2; ++j) {
          f[i][j].assign(n + 1, n + 1);
          trace[i][j].resize(n + 1);
        }
      f[0][0][0] = -n - 1;
      f[0][1][0] = -n - 1;
      for (int i = 0; i < n - 1; ++i) {
        if (f[0][0][i] <= n) {
          if (a[i + 1] > a[i])
            f[0][0][i + 1] = f[0][0][i], trace[0][0][i + 1] = {0, 0};
          else if (a[i + 1] > f[0][0][i])
            f[1][0][i + 1] = a[i], trace[1][0][i + 1] = {0, 0};
          if (-a[i + 1] > f[0][0][i])
            f[1][1][i + 1] = a[i], trace[1][1][i + 1] = {0, 0};
        }
        if (f[0][1][i] <= n) {
          if (f[0][1][i] < f[0][0][i + 1])
            f[0][0][i + 1] = f[0][1][i], trace[0][0][i + 1] = {0, 1};
          if (a[i + 1] < a[i])
            f[0][1][i + 1] = f[0][1][i], trace[0][1][i + 1] = {0, 1};
          else if (-a[i + 1] > f[0][1][i])
            f[1][1][i + 1] = -a[i], trace[1][1][i + 1] = {0, 1};
        }
        if (f[1][0][i] <= n) {
          if (a[i + 1] > f[1][0][i] && f[0][0][i + 1] > a[i])
            f[0][0][i + 1] = a[i], trace[0][0][i + 1] = {1, 0};
          else if (a[i + 1] > a[i] && f[1][0][i + 1] > f[1][0][i])
            f[1][0][i + 1] = f[1][0][i], trace[1][0][i + 1] = {1, 0};
        }
        if (f[1][1][i] <= n) {
          if (a[i + 1] > f[1][1][i] && f[0][0][i + 1] > -a[i])
            f[0][0][i + 1] = -a[i], trace[0][0][i + 1] = {1, 1};
          else if (f[1][0][i + 1] > f[1][1][i])
            f[1][0][i + 1] = f[1][1][i], trace[1][0][i + 1] = {1, 1};
          if (-a[i + 1] > f[1][1][i] && f[0][1][i + 1] > -a[i])
            f[0][1][i + 1] = -a[i], trace[0][1][i + 1] = {1, 1};
          else if (a[i + 1] < a[i] && f[1][1][i + 1] > f[1][1][i])
            f[1][1][i + 1] = f[1][1][i], trace[1][1][i + 1] = {1, 1};
        }
      }
      pair<int, int> res{-1, -1};
      for (int i = 0; i < 2; ++i)
        for (int j = 0; j < 2; ++j)
          if (f[i][j][n - 1] <= n) res = {i, j};
      if (res.first == -1) {
        cout << "NO" << endl;
        continue;
      }
      cout << "YES" << endl;
      vector<int> ans;
      for (int i = n - 1; i >= 0; --i) {
        ans.push_back(res.second ? -a[i] : a[i]);
        res = trace[res.first][res.second][i];
      }
      for (auto it = ans.rbegin(); it != ans.rend(); ++it) cout << *it << " ";
      cout << endl;
    }
    return 0;
  }
  