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;
  template <class T>
  bool chmax(T &a, const T &b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  bool chmin(T &a, const T &b) {
    if (a > b) {
      a = b;
      return 1;
    }
    return 0;
  }
  int T;
  int N;
  int P[1010101];
  pair<int, int> from[1010101][2][2];
  int dp[1010101][2][2];
  void solve() {
    int i, j, k, l, r, x, y;
    string s;
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &N);
      for (i = 0; i < (N); i++) scanf("%d", &P[i]);
      for (i = 0; i < (N); i++)
        dp[i][0][0] = dp[i][0][1] = dp[i][1][0] = dp[i][1][1] = 1LL << 30;
      dp[0][0][0] = dp[0][0][1] = -1LL << 30;
      int pos, sig, ns;
      for (i = 0; i < (N - 1); i++)
        for (pos = 0; pos < (2); pos++)
          for (sig = 0; sig < (2); sig++)
            if (dp[i][pos][sig] < 1 << 30) {
              if (pos == 0) {
                x = (sig ? -P[i] : P[i]);
                y = dp[i][pos][sig];
              } else {
                x = dp[i][pos][sig];
                y = (sig ? -P[i] : P[i]);
              }
              for (ns = 0; ns < (2); ns++) {
                int v = ns ? -P[i + 1] : P[i + 1];
                if (v > x) {
                  if (chmin(dp[i + 1][0][ns], y)) from[i + 1][0][ns] = {pos, sig};
                } else if (v > y) {
                  if (chmin(dp[i + 1][1][ns], x)) from[i + 1][1][ns] = {pos, sig};
                }
              }
            }
      pos = sig = -1;
      for (x = 0; x < (2); x++)
        for (y = 0; y < (2); y++)
          if (dp[N - 1][x][y] < 1 << 30) pos = x, sig = y;
      if (pos == -1) {
        cout << "NO" << endl;
      } else {
        for (i = N - 1; i >= 0; i--) {
          if (sig) P[i] = -P[i];
          auto p = from[i][pos][sig];
          pos = p.first;
          sig = p.second;
        }
        cout << "YES" << endl;
        for (i = 0; i < (N); i++) cout << P[i] << " ";
        cout << endl;
      }
    }
  }
  int main(int argc, char **argv) {
    string s;
    int i;
    if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
    for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
    for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
    cout.tie(0);
    solve();
    return 0;
  }
  