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;
  const int maxn = 1e6 + 2;
  const int INF = 1e9;
  int dp[maxn][2];
  int pre[maxn][2];
  int p[maxn];
  int a[maxn];
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> p[i];
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 2; j++) {
        dp[i][j] = INF;
      }
    }
    dp[0][0] = -INF;
    dp[0][1] = -INF;
    for (int i = 0; i < n - 1; i++) {
      for (int j = 0; j < 2; j++) {
        if (dp[i][j] != INF) {
          int x = j ? -p[i] : p[i];
          int y = dp[i][j];
          if (x < y) swap(x, y);
          for (int k = 0; k < 2; k++) {
            int z = k ? -p[i + 1] : p[i + 1];
            if (z > x) {
              if (dp[i + 1][k] > y) {
                dp[i + 1][k] = y;
                pre[i + 1][k] = j;
              }
            } else if (z > y) {
              if (dp[i + 1][k] > x) {
                dp[i + 1][k] = x;
                pre[i + 1][k] = j;
              }
            }
          }
        }
      }
    }
    for (int j = 0; j < 2; j++) {
      if (dp[n - 1][j] != INF) {
        for (int i = n - 1; i >= 0; i--) {
          a[i] = j ? -p[i] : p[i];
          j = pre[i][j];
        }
        cout << "YES\n";
        for (int i = 0; i < n; i++) {
          cout << a[i] << ' ';
        }
        cout << '\n';
        return;
      }
    }
    cout << "NO\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      solve();
    }
  }
  