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 a[1000000 + 5];
  int f[1000000 + 5][2][2];
  pair<int, int> g[1000000 + 5][2][2];
  void print(int n, int i, int j) {
    if (n == 0) return;
    print(n - 1, g[n][i][j].first, g[n][i][j].second);
    if (j == 0)
      printf("%d ", a[n]);
    else
      printf("%d ", -a[n]);
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      int n, i, j, k, l;
      scanf("%d", &n);
      for (i = 1; i <= n; i++) scanf("%d", &a[i]);
      for (i = 0; i <= n + 1; i++)
        for (j = 0; j < 2; j++)
          for (k = 0; k < 2; k++) f[i][j][k] = 1e9;
      f[1][0][0] = f[1][0][1] = -1e9;
      for (i = 1; i < n; i++)
        for (j = 0; j < 2; j++)
          for (k = 0; k < 2; k++) {
            int x = a[i], y = f[i][j][k];
            if (k) x = -x;
            if (j) swap(x, y);
            for (l = 0; l < 2; l++) {
              int z = a[i + 1];
              if (l) z = -z;
              if (z < y) continue;
              if (z < x) {
                if (f[i + 1][1][l] > x)
                  f[i + 1][1][l] = x, g[i + 1][1][l] = make_pair(j, k);
              } else {
                if (f[i + 1][0][l] > y)
                  f[i + 1][0][l] = y, g[i + 1][0][l] = make_pair(j, k);
              }
            }
          }
      bool flag = false;
      for (i = 0; i < 2 && !flag; i++)
        for (j = 0; j < 2 && !flag; j++)
          if (f[n][i][j] <= n && !flag) {
            printf("YES\n");
            print(n, i, j);
            flag = true;
            break;
          }
      if (!flag) printf("NO");
      printf("\n");
    }
  }
  