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;
  using cd = complex<double>;
  mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
  const long long N = 1e6 + 5;
  long long a[N], n, d[N][2][2];
  pair<long long, long long> p[N][2][2];
  long long sg(long long x) {
    if (x == 0)
      return 1;
    else
      return -1;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long q;
    cin >> q;
    while (q--) {
      cin >> n;
      for (long long i = 1; i <= n; i++) {
        cin >> a[i];
        d[i][0][0] = d[i][0][1] = d[i][1][0] = d[i][1][1] = 2e18;
      }
      d[1][0][1] = d[1][0][0] = -2e18;
      for (long long i = 1; i < n; i++) {
        for (long long j = 0; j < 2; j++) {
          for (long long k = 0; k < 2; k++) {
            for (long long v = 0; v < 2; v++) {
              if (d[i][j][k] > 1e18) continue;
              long long x = j ? d[i][j][k] : a[i] * sg(k);
              long long y = j ? a[i] * sg(k) : d[i][j][k];
              long long z = a[i + 1] * sg(v);
              if (x < y) continue;
              if (z > x) {
                if (d[i + 1][0][v] > y) {
                  d[i + 1][0][v] = y;
                  p[i + 1][0][v] = {j, k};
                }
              } else if (z > y) {
                if (d[i + 1][1][v] > x) {
                  d[i + 1][1][v] = x;
                  p[i + 1][1][v] = {j, k};
                }
              }
            }
          }
        }
      }
      long long x = -1, y = -1;
      for (long long i = 0; i < 2; i++) {
        for (long long j = 0; j < 2; j++) {
          if (d[n][i][j] < 2e18) {
            x = i;
            y = j;
          }
        }
      }
      if (x == -1)
        cout << "NO" << endl;
      else {
        for (long long i = n; i >= 1; i--) {
          if (y) a[i] *= -1;
          pair<long long, long long> v = p[i][x][y];
          x = v.first;
          y = v.second;
        }
        cout << "YES" << endl;
        for (long long i = 1; i <= n; i++) {
          cout << a[i] << " ";
        }
        cout << endl;
      }
    }
  }
  