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;
  long long inf = 1e18;
  int __FAST_IO__ = []() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);
    std::cout.tie(0);
    return 0;
  }();
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int N;
      cin >> N;
      vector<int> v(N);
      for (int i = 0; i < (N); ++i) cin >> v[i];
      vector<vector<vector<int>>> dp(
          N, vector<vector<int>>(2, vector<int>(2, INT_MAX)));
      vector<vector<vector<pair<int, int>>>> pr(
          N,
          vector<vector<pair<int, int>>>(2, vector<pair<int, int>>(2, {-1, -1})));
      dp[0][1][0] = dp[0][1][1] = INT_MIN;
      for (int i = (1); i < (N); ++i) {
        for (int pos = 0; pos < (2); ++pos)
          for (int sign = 0; sign < (2); ++sign) {
            if (dp[i - 1][pos][sign] == INT_MAX) continue;
            int x = sign ? -v[i - 1] : v[i - 1];
            int y = dp[i - 1][pos][sign];
            if (!pos) swap(x, y);
            for (int nsign = 0; nsign < (2); ++nsign) {
              int z = nsign ? -v[i] : v[i];
              if (z > x) {
                if (dp[i][1][nsign] > y) {
                  dp[i][1][nsign] = y;
                  pr[i][1][nsign] = {pos, sign};
                }
              } else if (z > y) {
                if (dp[i][0][nsign] > x) {
                  dp[i][0][nsign] = x;
                  pr[i][0][nsign] = {pos, sign};
                }
              }
            }
          }
      }
      for (int pos = 0; pos < (2); ++pos)
        for (int sign = 0; sign < (2); ++sign) {
          if (dp[N - 1][pos][sign] != INT_MAX) {
            printf("YES\n");
            int p = pos, s = sign;
            vector<int> ans;
            for (int i = (N - 1); i >= 0; --i) {
              ans.push_back(s ? -v[i] : v[i]);
              auto next = pr[i][p][s];
              p = next.first, s = next.second;
            }
            for (int i = (ans.size() - 1); i >= 0; --i) printf("%d ", ans[i]);
            printf("\n");
            goto finish;
          }
        }
      printf("NO\n");
    finish:;
    }
    return 0;
  }
  