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 n;
  int p[1000005];
  int f[1000005][2][2];
  tuple<int, int, int> pre[1000005][2][2];
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> p[i];
    for (int i = 0; i <= n; i++)
      for (int k1 = 0; k1 <= 1; k1++)
        for (int k2 = 0; k2 <= 1; k2++) f[i][k1][k2] = 1e9;
    f[1][0][0] = f[1][1][0] = -1e9;
    for (int i = 1; i < n; i++) {
      for (int k1 = 0; k1 <= 1; k1++) {
        for (int k2 = 0; k2 <= 1; k2++) {
          int x, y, t = f[i][k1][k2];
          x = (2 * k1 - 1) * p[i], y = t;
          if (k2) swap(x, y);
          auto nw = make_tuple(i, k1, k2);
          for (int sgn = 0; sgn <= 1; sgn++) {
            int z = (2 * sgn - 1) * p[i + 1];
            if (z < y) continue;
            if (z < x && f[i + 1][sgn][1] > x)
              f[i + 1][sgn][1] = x, pre[i + 1][sgn][1] = nw;
            if (z >= x && f[i + 1][sgn][0] > y)
              f[i + 1][sgn][0] = y, pre[i + 1][sgn][0] = nw;
          }
        }
      }
    }
    for (int k1 = 0; k1 <= 1; k1++) {
      for (int k2 = 0; k2 <= 1; k2++) {
        if (f[n][k1][k2] != 1e9) {
          vector<int> ans;
          int i = n;
          while (i) {
            ans.push_back((2 * k1 - 1) * p[i]);
            tie(i, k1, k2) = pre[i][k1][k2];
          }
          reverse(ans.begin(), ans.end());
          cout << "YES\n";
          for (auto x : ans) cout << x << ' ';
          cout << '\n';
          return;
        }
      }
    }
    cout << "NO\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) {
      solve();
    }
    return 0;
  }
  