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 stala = 1e6 + 10;
  int dp[stala][2];
  int poprzedni[stala][2];
  int tab[stala];
  int res[stala];
  void f2(int ile, int index) {
    for (int i = ile; i >= 1; i--) {
      if (index == 1) {
        res[i] = tab[i];
      } else {
        res[i] = -tab[i];
      }
      index = poprzedni[i][index];
    }
    for (int i = 1; i <= ile; i++) {
      cout << res[i] << " ";
    }
    cout << "\n";
  }
  void czysc(int ile) {
    for (int i = 1; i <= ile; i++) {
      dp[i][0] = 0;
      dp[i][1] = 0;
      poprzedni[i][0] = 0;
      poprzedni[i][1] = 0;
      tab[i] = 0;
      res[i] = 0;
    }
  }
  void f() {
    int ile;
    cin >> ile;
    for (int i = 1; i <= ile; i++) {
      cin >> tab[i];
    }
    dp[1][0] = -1e9;
    dp[1][1] = -1e9;
    for (int i = 2; i <= ile; i++) {
      dp[i][0] = 1e9;
      dp[i][1] = 1e9;
      if (dp[i - 1][0] != 1e9) {
        if (tab[i - 1] > tab[i]) {
          dp[i][0] = min(dp[i][0], dp[i - 1][0]);
        }
        if (dp[i - 1][0] < tab[i] * (-1)) {
          dp[i][0] = min(dp[i][0], tab[i - 1] * (-1));
        }
        if (tab[i - 1] * (-1) < tab[i]) {
          dp[i][1] = min(dp[i][1], dp[i - 1][0]);
        }
        if (dp[i - 1][0] < tab[i]) {
          dp[i][1] = min(dp[i][1], tab[i - 1] * (-1));
        }
      }
      if (dp[i - 1][1] != 1e9) {
        if (tab[i - 1] < tab[i] * (-1)) {
          if (dp[i][0] > dp[i - 1][1]) {
            dp[i][0] = dp[i - 1][1];
            poprzedni[i][0] = 1;
          }
        }
        if (dp[i - 1][1] < tab[i] * (-1)) {
          if (dp[i][0] > tab[i - 1]) {
            dp[i][0] = tab[i - 1];
            poprzedni[i][0] = 1;
          }
        }
        if (tab[i - 1] < tab[i]) {
          if (dp[i][1] > dp[i - 1][1]) {
            dp[i][1] = dp[i - 1][1];
            poprzedni[i][1] = 1;
          }
        }
        if (dp[i - 1][1] < tab[i]) {
          if (dp[i][1] > tab[i - 1]) {
            dp[i][1] = tab[i - 1];
            poprzedni[i][1] = 1;
          }
        }
      }
    }
    if (dp[ile][1] != 1e9 || dp[ile][0] != 1e9) {
      cout << "YES"
           << "\n";
      if (dp[ile][1] != 1e9) {
        f2(ile, 1);
      } else {
        f2(ile, 0);
      }
    } else {
      cout << "NO"
           << "\n";
    }
    czysc(ile);
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int testy;
    cin >> testy;
    for (int i = 1; i <= testy; i++) {
      f();
    }
    return 0;
  }
  