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() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> a(n + 1, 0);
      for (int i = 1; i <= n; i++) cin >> a[i];
      const int inf = (1 << 30);
      vector<array<int, 2>> dp(n + 1, {inf, inf});
      vector<array<int, 2>> p(n + 1, {-1, -1});
      dp[1] = {-inf, -inf};
      auto upd = [&](int i, int j, int lim, int u) {
        if (lim < dp[i][j]) {
          dp[i][j] = lim;
          p[i][j] = u;
        }
      };
      for (int i = 1; i < n; i++) {
        for (int j = 0; j < 2; j++) {
          if (dp[i][j] == inf) continue;
          int x = (j ? -a[i] : a[i]);
          for (int k = 0; k < 2; k++) {
            int y = (k ? -a[i + 1] : a[i + 1]);
            if (x < y) upd(i + 1, k, dp[i][j], j);
            if (dp[i][j] < y) upd(i + 1, k, x, j);
          }
        }
      }
      if (dp[n][0] != inf) {
        int s = 0;
        cout << "YES\n";
        vector<int> res(n + 1, 0);
        for (int i = n; i >= 1; i--) {
          res[i] = (s ? -a[i] : a[i]);
          s = p[i][s];
        }
        for (int i = 1; i <= n; i++) cout << res[i] << " \n"[i == n];
      } else
        cout << "NO\n";
    }
    return 0;
  }
  