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;
  struct State {
    int st, nd, val, prv;
    State(int st_, int nd_, int val_, int prv_)
        : st(st_), nd(nd_), val(val_), prv(prv_) {}
    bool operator>(const State& other) const {
      if (st != other.st) {
        return st > other.st;
      }
      return nd > other.nd;
    }
    void update(int x) {
      if (x > st) {
        st = x;
      } else if (x > nd) {
        nd = x;
      }
    }
    string to_string() const {
      stringstream ss;
      ss << st << ' ' << nd << ' ' << val << ' ' << prv;
      return ss.str();
    }
  };
  vector<State> cleanup(const vector<State>& v) {
    vector<State> cleaned;
    int n = v.size();
    vector<int> o(n);
    iota(o.begin(), o.end(), 0);
    sort(o.begin(), o.end(), [&v](int i1, int i2) { return v[i1] > v[i2]; });
    for (int i : o) {
      const State& state = v[i];
      while (!cleaned.empty() && (cleaned.back().nd >= state.nd)) {
        cleaned.pop_back();
      }
      cleaned.push_back(state);
    }
    return cleaned;
  }
  vector<int> solve(const vector<int>& p) {
    int n = p.size();
    int inf = n + 5;
    vector<vector<State>> dp(n);
    dp[0].emplace_back(p[0], -inf, p[0], -1);
    dp[0].emplace_back(-p[0], -inf, -p[0], -1);
    dp[0] = cleanup(dp[0]);
    for (int i = 1; i < n; ++i) {
      for (int j = 0; j < int(dp[i - 1].size()); ++j) {
        const State& prv = dp[i - 1][j];
        for (int flip = -1; flip < 2; flip += 2) {
          int val = p[i] * flip;
          if (val > prv.nd) {
            State new_state = prv;
            new_state.val = val;
            new_state.prv = j;
            new_state.update(val);
            dp[i].push_back(new_state);
          }
        }
      }
      dp[i] = cleanup(dp[i]);
    }
    if (dp[n - 1].empty()) {
      return {};
    }
    vector<int> a(n);
    int cur = 0;
    for (int i = n - 1; i >= 0; --i) {
      a[i] = dp[i][cur].val;
      cur = dp[i][cur].prv;
    }
    return a;
  }
  void test_case() {
    int n;
    cin >> n;
    vector<int> p(n);
    for (int i = 0; i < n; ++i) {
      cin >> p[i];
    }
    vector<int> ans = solve(p);
    if (ans.empty()) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      for (int i = 0; i < n; ++i) {
        if (i > 0) cout << ' ';
        cout << ans[i];
      }
      cout << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) {
      test_case();
    }
  }
  