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 MAX_N = 1e6 + 7;
  int p[MAX_N][2];
  int dp[MAX_N][2][2];
  pair<int, int> f[MAX_N][2][2];
  bool g[MAX_N][2][2];
  vector<int> ans;
  int T, n;
  void solve(int s, int k) {
    for (int i = n; i; --i) {
      ans.push_back(p[i][s]);
      pair<int, int> p = f[i][s][k];
      s = p.first, k = p.second;
    }
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; ++i) scanf("%d", &p[i][0]), p[i][1] = -p[i][0];
      for (int i = 1; i <= n; ++i) {
        for (int s = 0; s < 2; ++s) {
          for (int k = 0; k < 2; ++k) {
            dp[i][s][k] = -(1e9 + 7);
            g[i][s][k] = false;
          }
        }
      }
      g[1][0][0] = g[1][1][0] = true;
      for (int i = 2; i <= n; ++i) {
        for (int s1 = 0; s1 < 2; ++s1) {
          for (int s2 = 0; s2 < 2; ++s2) {
            if (p[i - 1][s1] < p[i][s2]) {
              if (g[i - 1][s1][1]) {
                if (p[i][s2] > dp[i - 1][s1][1] &&
                    (!g[i][s2][0] || dp[i][s2][0] > p[i - 1][s1])) {
                  dp[i][s2][0] = p[i - 1][s1];
                  f[i][s2][0] = pair<int, int>(s1, 1);
                  g[i][s2][0] = true;
                }
                if (p[i][s2] < dp[i - 1][s1][1] && p[i][s2] > p[i - 1][s1] &&
                    (!g[i][s2][1] || dp[i][s2][1] > dp[i - 1][s1][1])) {
                  dp[i][s2][1] = dp[i - 1][s1][1];
                  f[i][s2][1] = pair<int, int>(s1, 1);
                  g[i][s2][1] = true;
                }
              }
              if (g[i - 1][s1][0]) {
                if (dp[i][s2][0] > dp[i - 1][s1][0] || (!g[i][s2][0])) {
                  dp[i][s2][0] = dp[i - 1][s1][0];
                  g[i][s2][0] = true;
                  f[i][s2][0] = pair<int, int>(s1, 0);
                }
              }
            } else {
              if (g[i - 1][s1][0]) {
                if (p[i][s2] > dp[i - 1][s1][0] &&
                    (!g[i][s2][1] || dp[i][s2][1] > p[i - 1][s1])) {
                  dp[i][s2][1] = p[i - 1][s1];
                  f[i][s2][1] = pair<int, int>(s1, 0);
                  g[i][s2][1] = true;
                }
              }
            }
          }
        }
      }
      bool flag = true;
      for (int s = 0; s < 2; ++s) {
        for (int k = 0; k < 2; ++k) {
          if (g[n][s][k]) {
            solve(s, k);
            flag = false;
            break;
          }
        }
        if (!flag) break;
      }
      if (flag) {
        puts("NO");
      } else {
        puts("YES");
        for (int i = ans.size() - 1; i >= 0; --i) printf("%d ", ans[i]);
        puts("");
      }
      ans.clear();
    }
    return 0;
  }
  