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 it[1000001];
  int dp[1000001][2];
  pair<int, int> ba[1000001][2];
  int ma[2];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      int st = 1;
      ma[0] = -1;
      ma[1] = 1;
      for (int i = 0; i < n; i++) {
        cin >> it[i];
        if (it[i] != (i + 1)) {
          st = 0;
        }
      }
      if (n <= 2 or st == 1) {
        cout << "YES" << '\n';
        for (int i = 0; i < n; i++) {
          cout << it[i] << " ";
        }
        cout << '\n';
        continue;
      }
      int xx = -(n + 1);
      dp[0][0] = xx;
      dp[0][1] = xx;
      ba[0][0] = {-1, -1};
      ba[0][1] = {-1, -1};
      for (int i = 1; i < n; i++) {
        dp[i][0] = n + 1;
        dp[i][1] = n + 1;
        for (int j = 0; j < 2; j++) {
          for (int k = 0; k < 2; k++) {
            if (dp[i - 1][j] == n + 1) {
              continue;
            }
            if ((ma[k] * it[i]) > (ma[j] * it[i - 1])) {
              if (dp[i - 1][j] < dp[i][k]) {
                ba[i][k] = {i - 1, j};
              }
              dp[i][k] = min(dp[i][k], dp[i - 1][j]);
            }
            if ((it[i] * ma[k]) > dp[i - 1][j]) {
              if ((ma[j] * (it[i - 1])) < dp[i][k]) {
                ba[i][k] = {i - 1, j};
              }
              dp[i][k] = min(dp[i][k], (ma[j] * (it[i - 1])));
            }
          }
        }
      }
      if (dp[n - 1][0] == -xx and dp[n - 1][1] == -xx) {
        cout << "NO" << '\n';
      } else {
        pair<int, int> cur = {n - 1, 0};
        if (dp[n - 1][1] != -xx) {
          cur.second = 1;
        }
        cout << "YES" << '\n';
        vector<int> ans;
        while (cur.first >= 0) {
          ans.push_back(it[cur.first] * ma[cur.second]);
          cur = {ba[cur.first][cur.second].first,
                 ba[cur.first][cur.second].second};
        }
        for (int i = n - 1; i >= 0; i--) {
          cout << ans[i] << " ";
        }
        cout << '\n';
      }
    }
    return 0;
  }
  