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 = 1e6 + 7, inf = 1e9;
  int n, dp[N][2], w[N][2], lst[N][2], a[N], ns[N];
  void Main() {
    cin >> n;
    for (int i = (1); i <= (n); i++) cin >> a[i], w[i][0] = a[i], w[i][1] = -a[i];
    for (int i = (0); i <= (n); i++) dp[i][0] = dp[i][1] = inf;
    w[0][0] = w[0][1] = -inf;
    dp[0][0] = dp[0][1] = -inf;
    for (int i = (1); i <= (n); i++) {
      for (int a = (0); a <= (1); a++)
        for (int b = (0); b <= (1); b++)
          if (w[i][a] > w[i - 1][b] && dp[i][a] > dp[i - 1][b])
            dp[i][a] = dp[i - 1][b], lst[i][a] = b;
      for (int a = (0); a <= (1); a++)
        for (int b = (0); b <= (1); b++)
          if (w[i][a] > dp[i - 1][b] && dp[i][a] > w[i - 1][b])
            dp[i][a] = w[i - 1][b], lst[i][a] = b;
    }
    if (dp[n][0] == inf && dp[n][1] == inf)
      cout << "NO\n";
    else {
      cout << "YES\n";
      int o = dp[n][0] == inf ? 1 : 0;
      for (int i = (n); i >= (1); i--) ns[i] = w[i][o], o = lst[i][o];
      for (int i = (1); i <= (n); i++) cout << ns[i] << ' ';
      cout << '\n';
    }
  }
  int main() {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) Main();
    return 0;
  }
  