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 = 1e6 + 5;
  int n, p[N][2], dp[N][2][2];
  pair<int, int> g[N][2][2];
  void cmin(int& x, int y) {
    if (x > y) x = y;
  }
  void solve() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i][1]), p[i][0] = -p[i][1];
    for (int i = 1; i <= n; i++)
      for (int j = 0; j <= 1; j++)
        for (int k = 0; k <= 1; k++) dp[i][j][k] = n + 1;
    dp[1][0][0] = dp[1][0][1] = -n - 1;
    for (int i = 1; i < n; i++)
      for (int j = 0; j <= 1; j++)
        for (int k = 0; k <= 1; k++)
          for (int l = 0; l <= 1; l++) {
            int x = (j == 0) ? p[i][k] : dp[i][j][k],
                y = (j == 1) ? p[i][k] : dp[i][j][k];
            if (x > p[i + 1][l] && p[i + 1][l] > y && dp[i + 1][1][l] > x)
              dp[i + 1][1][l] = x, g[i + 1][1][l] = {j, k};
            if (p[i + 1][l] > x && dp[i + 1][0][l] > y)
              dp[i + 1][0][l] = y, g[i + 1][0][l] = {j, k};
          }
    vector<int> ans;
    for (int i = 0; i <= 1; i++)
      for (int j = 0; j <= 1; j++)
        if (dp[n][i][j] != n + 1) {
          puts("YES");
          pair<int, int> now = g[n][i][j];
          int tmp = n - 1;
          ans.push_back(p[n][j]);
          while (tmp) {
            ans.push_back(p[tmp][now.second]);
            now = g[tmp][now.first][now.second];
            tmp--;
          }
          for (int i = ans.size() - 1; i >= 0; i--) printf("%d ", ans[i]);
          puts("");
          return;
        }
    puts("NO");
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) solve();
  }
  