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 p[(1000000 + 1)], a[(1000000 + 1)], dp[(1000000 + 1)][2],
      pr[(1000000 + 1)][2];
  void solve() {
    long long n;
    cin >> n;
    for (long long i = 0; i < n; i++) {
      cin >> p[i];
      for (long long sign = 0; sign < 2; sign++) {
        dp[i][sign] = LONG_MAX;
      }
    }
    dp[0][0] = LONG_MIN;
    dp[0][1] = LONG_MIN;
    for (long long i = 0; i < n - 1; i++) {
      for (long long sign = 0; sign < 2; sign++) {
        long long x = sign ? -p[i] : p[i];
        long long y = dp[i][sign];
        if (x < y) swap(x, y);
        for (long long new_sign = 0; new_sign < 2; new_sign++) {
          long long z = new_sign ? -p[i + 1] : p[i + 1];
          if (z > x) {
            if (dp[i + 1][new_sign] > y) {
              dp[i + 1][new_sign] = y;
              pr[i + 1][new_sign] = sign;
            }
          } else if (z > y) {
            if (dp[i + 1][new_sign] > x) {
              dp[i + 1][new_sign] = x;
              pr[i + 1][new_sign] = sign;
            }
          }
        }
      }
    }
    long long sign = -1;
    for (long long new_sign = 0; new_sign < 2; new_sign++) {
      if (dp[n - 1][new_sign] != LONG_MAX) {
        sign = new_sign;
      }
    }
    if (sign == -1) {
      cout << "NO" << endl;
      return;
    } else {
      cout << "YES" << endl;
    }
    for (long long i = n - 1; i >= 0; i--) {
      a[i] = sign ? -p[i] : p[i];
      sign = pr[i][sign];
    }
    for (long long i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  