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;
  template <typename T>
  istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x : v) in >> x;
    return in;
  }
  template <typename T>
  ostream &operator<<(ostream &out, vector<T> &v) {
    for (auto &x : v) out << x << " ";
    return out;
  }
  template <typename Iterable>
  void prnv(const Iterable &iterable, ostream &out = cout) {
    if (iterable.begin() == iterable.end()) {
      out << '\n';
      return;
    }
    auto x = iterable.begin();
    out << *x;
    for (++x; x != iterable.end(); ++x) out << ' ' << *x;
    out << '\n';
  }
  void prn(char en = '\n') { cout << en; }
  template <typename T, typename... Args>
  void prn(T x, Args... args) {
    cout << x << " ";
    prn(args...);
  }
  void dbg() { cerr << '\n'; }
  template <typename T, typename... Args>
  void dbg(T x, Args... args) {}
  const int N = (int)1e6 + 5;
  int dp[2][N], v[2][N];
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> v[0][i];
      v[1][i] = -v[0][i];
    }
    dp[0][0] = dp[1][0] = -N;
    for (int i = 1; i < n; ++i) {
      for (int j = 0; j < 2; ++j) {
        dp[j][i] = N;
        for (int k = 0; k < 2; ++k) {
          if (v[j][i] >= v[k][i - 1]) {
            dp[j][i] = min(dp[j][i], dp[k][i - 1]);
          }
          if (v[j][i] >= dp[k][i - 1]) {
            dp[j][i] = min(dp[j][i], v[k][i - 1]);
          }
        }
      }
    }
    if (dp[0][n - 1] >= N && dp[1][n - 1] >= N) {
      cout << "NO\n";
      return;
    }
    cout << "YES\n";
    vector<int> ret;
    int j = 0;
    if (dp[j][n - 1] >= N) j = 1;
    ret.push_back(v[j][n - 1]);
    for (int i = n - 1; i > 0; --i) {
      if (v[j][i] >= v[j ^ 1][i - 1] && dp[j][i] == dp[j ^ 1][i - 1]) {
        j ^= 1;
      }
      if (v[j][i] >= dp[j ^ 1][i - 1] && dp[j][i] == v[j ^ 1][i - 1]) {
        j ^= 1;
      }
      ret.push_back(v[j][i - 1]);
    }
    reverse(ret.begin(), ret.end());
    cout << ret << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(10);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  