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;
  void solve() {
    int n;
    cin >> n;
    vector<vector<int>> a(n + 1, vector<int>(2));
    for (int i = 1; i <= n; ++i) cin >> a[i][0], a[i][1] = -a[i][0];
    vector<vector<vector<int>>> f(
        n + 1, vector<vector<int>>(2, vector<int>(2, INT_MAX)));
    vector<vector<vector<pair<int, int>>>> prv(
        n + 1,
        vector<vector<pair<int, int>>>(2, vector<pair<int, int>>(2, {0, 0})));
    f[1][0][0] = f[1][1][0] = INT_MIN;
    for (int i = 1; i < n; ++i) {
      for (int j = 0; j < 2; ++j) {
        for (int k = 0; k < 2; ++k)
          if (f[i][j][k] != INT_MAX) {
            for (int l = 0; l < 2; ++l) {
              if (k == 0) {
                if (a[i + 1][l] > a[i][j]) {
                  if (f[i][j][k] < f[i + 1][l][0]) {
                    f[i + 1][l][0] = f[i][j][k];
                    prv[i + 1][l][0] = {j, k};
                  }
                } else if (a[i + 1][l] > f[i][j][k]) {
                  if (a[i][j] < f[i + 1][l][1]) {
                    f[i + 1][l][1] = a[i][j];
                    prv[i + 1][l][1] = {j, k};
                  }
                }
              } else {
                if (a[i + 1][l] > f[i][j][k]) {
                  if (a[i][j] < f[i + 1][l][0]) {
                    f[i + 1][l][0] = a[i][j];
                    prv[i + 1][l][0] = {j, k};
                  }
                } else if (a[i + 1][l] > a[i][j]) {
                  if (f[i][j][k] < f[i + 1][l][1]) {
                    f[i + 1][l][1] = f[i][j][k];
                    prv[i + 1][l][1] = {j, k};
                  }
                }
              }
            }
          }
      }
    }
    for (int j = 0; j < 2; ++j) {
      for (int k = 0; k < 2; ++k) {
        if (f[n][j][k] != INT_MAX) {
          vector<int> ans(n + 1);
          for (int i = n; i > 0; --i) {
            ans[i] = a[i][j];
            pair<int, int> t = prv[i][j][k];
            j = t.first, k = t.second;
          }
          cout << "YES\n";
          for (int i = 1; i <= n; ++i) cout << ans[i] << " \n"[i == n];
          return;
        }
      }
    }
    cout << "NO\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  