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>
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  using ll = long long;
  using ii = pair<int, int>;
  using ull = unsigned long long;
  using db = long double;
  const int N = 1e6 + 5, LG = 18, MOD = 998244353;
  const long double PI = acos(-1);
  int dp[N][2];
  int par[N][2];
  void doWork() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      for (int j = 0; j < 2; j++) dp[i][j] = INT_MAX;
    }
    dp[0][0] = dp[0][1] = INT_MIN;
    auto upd = [&](int i, int j, int k, int p) {
      if (dp[i][j] > k) {
        dp[i][j] = k;
        par[i][j] = p;
      }
    };
    for (int i = 0; i < n - 1; i++)
      for (int x = 0; x < 2; x++)
        for (int y = 0; y < 2; y++) {
          int val1 = (x ? -1 : 1) * a[i];
          int val2 = (y ? -1 : 1) * a[i + 1];
          if (val1 <= val2) {
            upd(i + 1, y, dp[i][x], x);
          }
          if (dp[i][x] <= val2) {
            upd(i + 1, y, val1, x);
          }
        }
    if (dp[n - 1][0] == INT_MAX) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      int cur = 0;
      for (int i = n - 1; i >= 0; --i) {
        if (cur) a[i] *= -1;
        cur = par[i][cur];
      }
      for (int i = 0; i < n; i++) cout << a[i] << " \n"[i + 1 == n];
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
      doWork();
    }
    return 0;
  }
  