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;
  int n, p[1000100];
  int dp[1000100][2][2];
  pair<bool, bool> pre[1000100][2][2];
  void dfs(int u, int x, int y) {
    if (u == n + 1) return;
    cout << (!x ? -p[u] : p[u]) << ' ';
    dfs(u + 1, pre[u][x][y].first, pre[u][x][y].second);
  }
  void Solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cin >> p[i];
      dp[i][0][0] = dp[i][0][1] = dp[i][1][0] = dp[i][1][1] = -n - 1;
    }
    dp[n][0][0] = dp[n][1][0] = n + 1;
    for (int i = n - 1; i >= 1; --i) {
      for (int c = 0; c < 2; ++c) {
        int z = !c ? -p[i] : p[i];
        if (dp[i + 1][0][0] >= -n) {
          int x = -p[i + 1], y = dp[i + 1][0][0];
          if (z < x) {
            if (y > dp[i][c][0]) {
              dp[i][c][0] = y, pre[i][c][0] = pair<bool, bool>(0, 0);
            }
          } else if (z < y) {
            if (x > dp[i][c][1]) {
              dp[i][c][1] = x, pre[i][c][1] = pair<bool, bool>(0, 0);
            }
          }
        }
        if (dp[i + 1][1][0] >= -n) {
          int x = p[i + 1], y = dp[i + 1][1][0];
          if (z < x) {
            if (y > dp[i][c][0]) {
              dp[i][c][0] = y, pre[i][c][0] = pair<bool, bool>(1, 0);
            }
          } else if (z < y) {
            if (x > dp[i][c][1]) {
              dp[i][c][1] = x, pre[i][c][1] = pair<bool, bool>(1, 0);
            }
          }
        }
        if (dp[i + 1][0][1] >= -n) {
          int y = -p[i + 1], x = dp[i + 1][0][1];
          if (z < x) {
            if (y > dp[i][c][0]) {
              dp[i][c][0] = y, pre[i][c][0] = pair<bool, bool>(0, 1);
            }
          } else if (z < y) {
            if (x > dp[i][c][1]) {
              dp[i][c][1] = x, pre[i][c][1] = pair<bool, bool>(0, 1);
            }
          }
        }
        if (dp[i + 1][1][1] >= -n) {
          int y = p[i + 1], x = dp[i + 1][1][1];
          if (z < x) {
            if (y > dp[i][c][0]) {
              dp[i][c][0] = y, pre[i][c][0] = pair<bool, bool>(1, 1);
            }
          } else if (z < y) {
            if (x > dp[i][c][1]) {
              dp[i][c][1] = x, pre[i][c][1] = pair<bool, bool>(1, 1);
            }
          }
        }
      }
    }
    for (int i = 0; i < 2; ++i) {
      for (int j = 0; j < 2; ++j) {
        if (dp[1][i][j] >= -n) {
          cout << "YES\n";
          dfs(1, i, j);
          cout << '\n';
          return;
        }
      }
    }
    cout << "NO\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) Solve();
    return 0;
  }
  