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;
  using ll = long long;
  using ld = long double;
  template <class T>
  bool uin(T& a, const T& b) {
    return a > b ? a = b, true : false;
  }
  template <class T>
  bool uax(T& a, const T& b) {
    return a < b ? a = b, true : false;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  constexpr int inf = 1e9;
  void solve() {
    int n;
    cin >> n;
    vector<int> p(n);
    for (auto& x : p) cin >> x;
    p.insert(begin(p), -inf);
    vector<array<int, 2>> dp(n + 1);
    vector<array<bool, 2>> par(n + 1);
    fill(begin(dp[0]), end(dp[0]), -inf);
    for (int i = 1; i <= n; ++i) {
      auto& nxt = dp[i];
      auto& pre = dp[i - 1];
      fill(begin(nxt), end(nxt), inf);
      for (int x = -p[i], j = 0; j < 2; ++j, x = -x) {
        if (x > -p[i - 1] && uin(nxt[j], pre[0])) par[i][j] = 0;
        if (x > pre[0] && uin(nxt[j], -p[i - 1])) par[i][j] = 0;
        if (x > +p[i - 1] && uin(nxt[j], pre[1])) par[i][j] = 1;
        if (x > pre[1] && uin(nxt[j], +p[i - 1])) par[i][j] = 1;
      }
    }
    if (min(dp[n][0], dp[n][1]) > n) {
      cout << "NO\n";
      return;
    }
    cout << "YES\n";
    vector<int> res;
    for (int i = n, state = dp[n][0] <= n ? 0 : 1; i > 0; --i) {
      if (state)
        res.push_back(+p[i]);
      else
        res.push_back(-p[i]);
      state = par[i][state];
    }
    reverse(begin(res), end(res));
    for (auto x : res) cout << x << ' ';
    cout << '\n';
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) solve();
  }
  