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 + 5, inf = 1e9;
  int T, n, p[N], dp[N][2][2], a[N];
  pair<int, int> pre[N][2][2];
  void print() {
    for (int pos = 0; pos < 2; pos++)
      for (int sg = 0; sg < 2; sg++)
        if (dp[n][pos][sg] != inf) {
          printf("YES\n");
          for (int j = n; j >= 1; j--) {
            a[j] = sg ? p[j] : -p[j];
            auto pr = pre[j][pos][sg];
            pos = pr.first, sg = pr.second;
          }
          for (int j = 1; j <= n; j++) printf("%d ", a[j]);
          printf("\n");
          return;
        }
    printf("NO\n");
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
      for (int i = 1; i <= n; i++)
        for (int pos = 0; pos < 2; pos++)
          for (int sg = 0; sg < 2; sg++) dp[i][pos][sg] = inf;
      dp[1][0][0] = -inf, dp[1][0][1] = -inf;
      for (int i = 1; i < n; i++)
        for (int pos = 0; pos < 2; pos++)
          for (int sg = 0; sg < 2; sg++)
            if (dp[i][pos][sg] != inf) {
              for (int sig = 0; sig < 2; sig++) {
                int z = sig ? p[i + 1] : -p[i + 1], x, y;
                if (!pos)
                  x = sg ? p[i] : -p[i], y = dp[i][pos][sg];
                else
                  x = dp[i][pos][sg], y = sg ? p[i] : -p[i];
                if (y < z && z < x) {
                  if (x < dp[i + 1][1][sig]) {
                    dp[i + 1][1][sig] = x;
                    pre[i + 1][1][sig] = {pos, sg};
                  }
                } else if (z > x) {
                  if (y < dp[i + 1][0][sig]) {
                    dp[i + 1][0][sig] = min(dp[i + 1][0][sig], y);
                    pre[i + 1][0][sig] = {pos, sg};
                  }
                }
              }
            }
      print();
    }
    return 0;
  }
  