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;
  constexpr int kN = 1e6 + 10;
  int dp[kN][2][2];
  pair<int, int> prefix[kN][2][2];
  void Solve() {
    constexpr int kInf = 1e9;
    int n;
    cin >> n;
    vector<int> p(n);
    for (int i = 0; i < n; i++) cin >> p[i];
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j < 2; j++) {
        for (int k = 0; k < 2; k++) dp[i][j][k] = kInf;
      }
    }
    dp[0][0][0] = dp[0][0][1] = -kInf;
    for (int i = 0; i < n - 1; i++) {
      for (int j = 0; j < 2; j++) {
        for (int k = 0; k < 2; k++)
          if (dp[i][j][k] != kInf) {
            int x = k == 0 ? p[i] : -p[i];
            int y = dp[i][j][k];
            if (x < y) swap(x, y);
            for (int o = 0; o < 2; o++) {
              int new_x = o == 0 ? p[i + 1] : -p[i + 1];
              if (new_x > x) {
                if (dp[i + 1][0][o] > y) {
                  dp[i + 1][0][o] = y;
                  prefix[i + 1][0][o] = make_pair(j, k);
                }
              } else if (new_x > y) {
                if (dp[i + 1][1][o] > x) {
                  dp[i + 1][1][o] = x;
                  prefix[i + 1][1][o] = make_pair(j, k);
                }
              }
            }
          }
      }
    }
    int nj = -1, nk = -1;
    for (int j = 0; j < 2; j++) {
      for (int k = 0; k < 2; k++) {
        if (dp[n - 1][j][k] != kInf) nj = j, nk = k;
      }
    }
    if (nj == -1 && nk == -1) {
      cout << "NO\n";
      return;
    }
    vector<int> a(n);
    for (int i = n - 1; i >= 0; i--) {
      a[i] = nk == 0 ? p[i] : -p[i];
      tie(nj, nk) = prefix[i][nj][nk];
    }
    cout << "YES\n";
    for (auto x : a) cout << x << ' ';
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T;
    cin >> T;
    while (T--) Solve();
    return 0;
  }
  