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>
  #pragma GCC optimize("Ofast,unroll-loops")
  #pragma GCC target("avx,avx2,sse,sse2")
  using namespace std;
  using ll = long long;
  template <typename F>
  void multitest(F func) {
    int t;
    cin >> t;
    while (t--) func();
  }
  void report(int ok) { cout << (ok ? "YES" : "NO") << '\n'; }
  const int N = 1e6 + 16;
  int dp[N][2];
  int p[N][2];
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    multitest([&]() {
      int n;
      cin >> n;
      vector<int> a(n);
      for (auto &i : a) cin >> i;
      for (int i = 0; i <= n; i++) dp[i][0] = dp[i][1] = 1 << 30;
      dp[0][0] = dp[0][1] = -(1 << 30);
      auto upd = [&](int i, int j, int k, int U) {
        if (dp[i][j] > k) {
          dp[i][j] = k;
          p[i][j] = U;
        }
      };
      for (int i = 0; i + 1 < n; i++) {
        for (int z = 0; z < 2; z++) {
          int v = z ? -a[i] : a[i];
          for (int f = 0; f < 2; f++) {
            int vv = f ? -a[i + 1] : a[i + 1];
            if (v <= vv) {
              upd(i + 1, f, dp[i][z], z);
            }
            if (dp[i][z] <= vv) {
              upd(i + 1, f, v, z);
            }
          }
        }
      }
      if (dp[n - 1][0] == (1 << 30)) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        int t = 0;
        for (int i = n - 1; i >= 0; i--) {
          if (t) a[i] *= -1;
          t = p[i][t];
        }
        for (auto i : a) cout << i << " ";
        cout << '\n';
      }
    });
  }
  