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 INF = 1e9;
  const int MOD = 1e9 + 7;
  const int N = 1e6 + 100;
  void tmax(int& a, int b) { a = max(a, b); }
  void tmin(int& a, int b) { a = min(a, b); }
  int a[N];
  int dp[N][2][2];
  pair<int, int> fat[N][2][2];
  int f(int a) {
    if (a == 1) return a;
    return a - 1;
  }
  void proc(int i, int b1, int b2, int b3, int b4) {
    if (dp[i - 1][b3][b4] == INF) return;
    if (b2 == b4) {
      if (a[i - 1] * f(b3) > a[i] * f(b1)) return;
      int adv;
      if (b2 == 1)
        adv = dp[i - 1][b3][b4];
      else
        adv = max(dp[i - 1][b3][b4], a[i] * f(b1));
      if (dp[i][b1][b2] > adv) {
        tmin(dp[i][b1][b2], adv);
        fat[i][b1][b2] = {b3, b4};
      }
    } else {
      if (dp[i - 1][b3][b4] > a[i] * f(b1)) return;
      int adv;
      if (b2 == 1)
        adv = a[i - 1] * f(b3);
      else
        adv = max(a[i - 1] * f(b3), a[i] * f(b1));
      if (dp[i][b1][b2] > adv) {
        tmin(dp[i][b1][b2], adv);
        fat[i][b1][b2] = {b3, b4};
      }
    }
  }
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      dp[i][0][0] = dp[i][0][1] = INF;
      dp[i][1][0] = dp[i][1][1] = INF;
    }
    cin >> a[0];
    dp[0][0][0] = -a[0];
    dp[0][1][0] = a[0];
    dp[0][0][1] = dp[0][1][1] = -INF;
    for (int i = 1; i < n; i++) {
      cin >> a[i];
      for (int b = 0; b < 16; b++) {
        int b1 = b & 1, b2 = (b >> 1) & 1, b3 = (b >> 2) & 1, b4 = (b >> 3) & 1;
        proc(i, b1, b2, b3, b4);
      }
    }
    if (dp[n - 1][1][0] == INF)
      cout << "NO" << '\n';
    else {
      vector<int> ans(n);
      int pos = n - 1, b1 = 1, b2 = 0;
      while (pos > 0) {
        ans[pos] = a[pos] * f(b1);
        int b3 = fat[pos][b1][b2].first;
        int b4 = fat[pos][b1][b2].second;
        b1 = b3;
        b2 = b4;
        pos--;
      }
      ans[pos] = a[pos] * f(b1);
      cout << "YES" << '\n';
      for (int x : ans) cout << x << " ";
      cout << '\n';
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    int tt;
    cin >> tt;
    while (tt--) solve();
    return 0;
  }
  