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 <class a, class b>
  ostream& operator<<(ostream& os, const pair<a, b>& p) {
    os << "(" << p.first << "," << p.second << ")";
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "[";
    if (!v.size())
      os << "]";
    else
      for (int i = 0; i < v.size(); ++i) os << v[i] << ",]"[i == v.size() - 1];
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const set<T>& s) {
    os << "{";
    if (!s.size())
      os << "}";
    else
      for (auto x : s) os << x << ",}"[x == *s.rbegin()];
    return os;
  }
  void printr(ostream& os) { os << '\n'; }
  template <class T, class... Args>
  void printr(ostream& os, const T& t, const Args&... args) {
    os << t;
    if (sizeof...(args)) os << " ";
    printr(os, args...);
  }
  template <class T>
  void read(T& x) {
    cin >> x;
  }
  template <class T, class... Args>
  void read(T& t, Args&... args) {
    read(t), read(args...);
  }
  template <class T>
  void readarray(T* A, int n) {
    for (__typeof(0) i = (0), _end = (n), _step = 1 - 2 * ((0) > (n)); i != _end;
         i += _step)
      read(A[i]);
  }
  void dbgr(ostream& os) { os << endl; }
  template <class T, class... Args>
  void dbgr(ostream& os, const T& t, const Args&... args) {
    os << t;
    if (sizeof...(args)) os << ", ";
    dbgr(os, args...);
  }
  long long randll(long long a, long long b) {
    uniform_int_distribution<int> u(a, b);
    static mt19937 e(time(0));
    return u(e);
  }
  long long pw(long long a, long long k = 998244353 - 2) {
    long long ans = 1;
    for (a %= 998244353; k; k >>= 1) {
      if (k & 1) ans = ans * a % 998244353;
      a = a * a % 998244353;
    }
    return ans;
  }
  int n, p[1000050];
  int dp[2][1000050][2];
  pair<int, int> prt[2][1000050][2];
  auto f = dp[0], g = dp[1];
  auto pf = prt[0], pg = prt[1];
  void chkmin(int& a, int b) { a = min(a, b); }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int _;
    read(_);
    while (_--) {
      cin >> n;
      for (int i = 1; i <= n; ++i) cin >> p[i];
      for (int i = 1; i <= n; ++i)
        for (int c = 0; c <= 1; ++c) f[i][c] = g[i][c] = 1000050;
      f[1][0] = f[1][1] = -1000050;
      for (int i = 1; i < n; ++i)
        for (int c = 0; c <= 1; ++c) {
          int a = c ? -p[i] : p[i];
          for (int nc = 0; nc <= 1; ++nc) {
            int na = nc ? -p[i + 1] : p[i + 1];
            if (f[i][c] < na && na < a && g[i + 1][nc] > a)
              g[i + 1][nc] = a, pg[i + 1][nc] = make_pair(0, c);
            if (a < na && f[i + 1][nc] > f[i][c])
              f[i + 1][nc] = f[i][c], pf[i + 1][nc] = make_pair(0, c);
            if (a < na && na < g[i][c] && g[i + 1][nc] > g[i][c])
              g[i + 1][nc] = g[i][c], pg[i + 1][nc] = make_pair(1, c);
            if (g[i][c] < na && f[i + 1][nc] > a)
              f[i + 1][nc] = a, pf[i + 1][nc] = make_pair(1, c);
          }
        }
      int k = 0, c = 0;
      for (int x = 0; x <= 1; ++x)
        for (int y = 0; y <= 1; ++y)
          if (dp[x][n][y] <= n) k = x, c = y;
      if (dp[k][n][c] > n)
        cout << "NO" << endl;
      else {
        cout << "YES" << endl;
        function<void(int, int, int)> pr = [&](int k, int i, int c) {
          if (i > 1) pr(prt[k][i][c].first, i - 1, prt[k][i][c].second);
          cout << (c ? -p[i] : p[i]) << " \n"[i == n];
        };
        pr(k, n, c);
      }
    }
    return 0;
  }
  