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>
  const int MAXN = 1000005;
  const long long INF = (long long)1e9 + 8763;
  using namespace std;
  template <class T>
  using min_heap = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  using max_heap = priority_queue<T, vector<T>, less<T>>;
  struct State {
    int j, k;
  };
  int n, P[MAXN];
  int dp[MAXN][2][2];
  State from[MAXN][2][2];
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> P[i];
      for (int j = 0; j <= 1; j++) {
        for (int k = 0; k <= 1; k++) {
          dp[i][j][k] = INF;
        }
      }
    }
    dp[1][0][0] = -INF;
    from[1][0][0] = {0, 0};
    dp[1][0][1] = -INF;
    from[1][0][1] = {0, 0};
    for (int i = 1; i < n; i++) {
      for (int j = 0; j <= 1; j++) {
        for (int k = 0; k <= 1; k++) {
          if (dp[i][j][k] == INF) continue;
          int x = P[i], y = dp[i][j][k];
          if (k) x *= -1;
          if (j) swap(x, y);
          assert(x > y);
          for (int t = 0; t <= 1; t++) {
            int nxt = (t ? -P[i + 1] : P[i + 1]);
            assert(nxt != x && nxt != y);
            if (nxt > x) {
              if (dp[i + 1][0][t] > y) {
                dp[i + 1][0][t] = y;
                from[i + 1][0][t] = {j, k};
              }
            } else if (nxt > y) {
              if (dp[i + 1][1][t] > x) {
                dp[i + 1][1][t] = x;
                from[i + 1][1][t] = {j, k};
              }
            }
          }
        }
      }
    }
    State ans = {-1, -1};
    for (int j = 0; j <= 1; j++) {
      for (int k = 0; k <= 1; k++) {
        if (dp[n][j][k] != INF) {
          ans = {j, k};
          break;
        }
      }
    }
    if (ans.j != -1) {
      cout << "YES\n";
      for (int i = n; i >= 1; i--) {
        if (ans.k) {
          P[i] *= -1;
        }
        ans = from[i][ans.j][ans.k];
      }
      for (int i = 1; i <= n; i++) {
        cout << P[i] << " \n"[i == n];
      }
    } else {
      cout << "NO\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int tc = 1;
    cin >> tc;
    for (int i = 1; i <= tc; i++) {
      solve();
    }
    return 0;
  }
  