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 i, j, k, n, m, t, it, a[1005000], fa[10005000];
  bool op[10050000];
  struct SB {
    int id, x, y;
    bool operator<(const SB a) const { return (x == a.x) ? y < a.y : x < a.x; }
  };
  vector<SB> v1, v2;
  void add(int x, int y, int z, int b) {
    fa[++it] = x;
    op[it] = b;
    v2.push_back({it, y, z});
  }
  int main() {
    ios::sync_with_stdio(0);
    cin >> t;
    while (t--) {
      cin >> n;
      it = 0;
      for (i = 1; i <= n; i++) cin >> a[i];
      v1.push_back({0, -11451419, -11451419});
      for (i = 1; i <= n; i++) {
        v2.clear();
        for (auto &[id, x, y] : v1) {
          if (x < -a[i])
            add(id, -a[i], y, 1);
          else if (y < -a[i])
            add(id, x, -a[i], 1);
          if (x < a[i])
            add(id, a[i], y, 0);
          else if (y < a[i])
            add(id, x, a[i], 0);
        }
        if (v2.empty()) {
          cout << "NO\n";
          goto ccc;
        }
        v1.clear();
        k = 1e9;
        sort(v2.begin(), v2.end());
        for (auto &i : v2) {
          if (i.y < k) {
            k = i.y;
            v1.push_back(i);
          }
        }
      }
      i = (*v1.begin()).id;
      j = n;
      while (i) {
        if (op[i]) a[j] *= -1;
        j--;
        i = fa[i];
      }
      cout << "YES\n";
      for (i = 1; i <= n; i++) cout << a[i] << ' ';
      cout << '\n';
    ccc:;
    }
  }
  