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;
  long long t, n, ind[2000005], v[1000005];
  pair<int, int> check(int val, pair<int, int> x, pair<int, int> y, int pos) {
    int precx = (pos - 2) / 2 * 2, precy = (pos - 2) / 2 * 2 + 1;
    int best = INT_MAX, bestind = -1;
    if (x.first < val && x.second < best) {
      best = x.second;
      bestind = precx;
    }
    if (x.second < val && x.first < best) {
      best = x.first;
      bestind = precx;
    }
    if (y.first < val && y.second < best) {
      best = y.second;
      bestind = precy;
    }
    if (y.second < val && y.first < best) {
      best = y.first;
      bestind = precy;
    }
    if (best == INT_MAX) {
      ind[pos] = -1;
      return {INT_MAX, INT_MAX};
    }
    if (pos > 1) ind[pos] = bestind;
    return {val, best};
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> t;
    while (t--) {
      cin >> n;
      for (int i = 0; i < 2 * n + 4; i++) ind[i] = -1;
      pair<int, int> x = {INT_MIN, INT_MIN}, y = {INT_MIN, INT_MIN};
      for (int i = 0; i < n; i++) {
        cin >> v[i];
        pair<int, int> succx = check(v[i], x, y, i * 2);
        pair<int, int> succy = check(-v[i], x, y, i * 2 + 1);
        x = succx;
        y = succy;
      }
      if (n == 1) {
        cout << "YES\n1\n";
        continue;
      }
      int st = -1;
      if (ind[2 * (n - 1) + 1] != -1) st = 2 * (n - 1) + 1;
      if (ind[2 * (n - 1)] != -1) st = 2 * (n - 1);
      if (st == -1)
        cout << "NO\n";
      else {
        cout << "YES\n";
        vector<int> ans;
        while (st >= 0) {
          if (st % 2 == 1)
            ans.push_back(v[st / 2] * (-1));
          else
            ans.push_back(v[st / 2]);
          st = ind[st];
        }
        reverse(ans.begin(), ans.end());
        for (auto i : ans) cout << i << " ";
        cout << "\n";
      }
    }
    return 0;
  }
  