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 N = 3e6 + 1e3 + 7;
  int T, n, a[N], vis[N];
  pair<int, int> f[N][2];
  int last[N][2];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> T;
    while (T--) {
      cin >> n;
      for (int i = 1; i <= n; i++) cin >> a[i], vis[i] = 0;
      for (int i = 1; i <= n; i++)
        for (int j = 0; j < 2; j++) f[i][j] = {1e9, 1e9};
      f[0][0] = f[0][1] = {-1e9, -1e9};
      for (int i = 1; i <= n; i++) {
        pair<int, int> dp1 = f[i - 1][0], dp2 = f[i - 1][1];
        if (-a[i] > dp1.first)
          dp1.first = -a[i];
        else if (-a[i] > dp1.second)
          dp1.second = -a[i];
        else
          dp1 = {1e9, 1e9};
        if (-a[i] > dp2.first)
          dp2.first = -a[i];
        else if (-a[i] > dp2.second)
          dp2.second = -a[i];
        else
          dp2 = {1e9, 1e9};
        if (dp1 < dp2)
          f[i][0] = dp1, last[i][0] = 0;
        else
          f[i][0] = dp2, last[i][0] = 1;
        dp1 = f[i - 1][0], dp2 = f[i - 1][1];
        if (a[i] > dp1.first)
          dp1.first = a[i];
        else if (a[i] > dp1.second)
          dp1.second = a[i];
        else
          dp1 = {1e9, 1e9};
        if (a[i] > dp2.first)
          dp2.first = a[i];
        else if (a[i] > dp2.second)
          dp2.second = a[i];
        else
          dp2 = {1e9, 1e9};
        if (dp1 < dp2)
          f[i][1] = dp1, last[i][1] = 0;
        else
          f[i][1] = dp2, last[i][1] = 1;
      }
      int ok = 0;
      if (f[n][0].first <= n && f[n][0].second <= n) {
        ok = 1;
        int q = 0;
        for (int i = n; i >= 1; i--) {
          if (!q) a[i] *= -1;
          q = last[i][q];
        }
      } else if (f[n][1].first <= n && f[n][1].second <= n) {
        ok = 1;
        int q = 1;
        for (int i = n; i >= 1; i--) {
          if (!q) a[i] *= -1;
          q = last[i][q];
        }
      }
      if (ok) {
        cout << "YES\n";
        for (int i = 1; i <= n; i++) cout << a[i] << " \n"[i == n];
      } else
        cout << "NO\n";
    }
  }
  