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() {
    long long n;
    cin >> n;
    vector<long long> v;
    vector<long long> ans(n, 0);
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      v.push_back(x);
    }
    vector<pair<long long, long long> > p[2];
    p[0].push_back(pair<long long, long long>(-1, LLONG_MIN));
    p[1].push_back(pair<long long, long long>(-1, LLONG_MIN));
    for (long long i = 1; i < n; i++) {
      long long x = v[i];
      long long y = -1 * v[i];
      pair<long long, long long> a = p[0][i - 1];
      pair<long long, long long> b = p[1][i - 1];
      pair<long long, long long> a0(2, LLONG_MAX);
      pair<long long, long long> a1(2, LLONG_MAX);
      if (a.first != 2) {
        if (v[i - 1] < x) {
          if (a0.second > a.second) {
            a0 = pair<long long, long long>(0, a.second);
          }
        }
        if (a.second < x) {
          if (a0.second > v[i - 1]) {
            a0 = pair<long long, long long>(0, v[i - 1]);
          }
        }
        if (v[i - 1] < y) {
          if (a1.second > a.second) {
            a1 = pair<long long, long long>(0, a.second);
          }
        }
        if (a.second < y) {
          if (a1.second > v[i - 1]) {
            a1 = pair<long long, long long>(0, v[i - 1]);
          }
        }
      }
      if (b.first != 2) {
        if (-1 * v[i - 1] < x) {
          if (a0.second > b.second) {
            a0 = pair<long long, long long>(1, b.second);
          }
        }
        if (b.second < x) {
          if (a0.second > -1 * v[i - 1]) {
            a0 = pair<long long, long long>(1, -1 * v[i - 1]);
          }
        }
        if (-1 * v[i - 1] < y) {
          if (a1.second > b.second) {
            a1 = pair<long long, long long>(1, b.second);
          }
        }
        if (b.second < y) {
          if (a1.second > -1 * v[i - 1]) {
            a1 = pair<long long, long long>(1, -1 * v[i - 1]);
          }
        }
      }
      p[0].push_back(a0);
      p[1].push_back(a1);
    }
    if (p[0][n - 1].first != 2) {
      ans[n - 1] = v[n - 1];
      long long c = 0;
      for (int i = n - 2; i >= 0; i--) {
        c = p[c][i + 1].first;
        if (c == 0) {
          ans[i] = v[i];
        } else {
          ans[i] = -1 * v[i];
        }
      }
    } else if (p[1][n - 1].first != 2) {
      ans[n - 1] = -1 * v[n - 1];
      long long c = 1;
      for (int i = n - 2; i >= 0; i--) {
        c = p[c][i + 1].first;
        if (c == 0) {
          ans[i] = v[i];
        } else {
          ans[i] = -1 * v[i];
        }
      }
    } else {
      cout << "NO\n";
      return;
    }
    cout << "YES\n";
    for (long long i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
  int main() {
    int n;
    cin >> n;
    while (n--) {
      solve();
    }
  }
  