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;
  int cas;
  int arr[2000005];
  int dp[2000005][2][2];
  int casio[2000005][2][2];
  int f(int i, int flag, int sgn) {
    if (i == 0) return -10000000;
    int cur = arr[i];
    if (sgn) cur *= -1;
    int ret = 10000000;
    if (casio[i][flag][sgn] == cas) return dp[i][flag][sgn];
    casio[i][flag][sgn] = cas;
    if (flag) {
      int pre = arr[i - 1];
      int a = f(i - 1, 0, 0);
      if (cur > pre && cur > a) ret = min(ret, pre);
      a = f(i - 1, 1, 0);
      if (cur > pre && cur > a) ret = min(ret, a);
      pre *= -1;
      a = f(i - 1, 0, 1);
      if (cur > pre && cur > a) ret = min(ret, pre);
      a = f(i - 1, 1, 1);
      if (cur > pre && cur > a) ret = min(ret, a);
    } else {
      int pre = arr[i - 1];
      int a = f(i - 1, 0, 0);
      if (cur > pre && cur < a) ret = min(ret, a);
      a = f(i - 1, 1, 0);
      if (cur < pre && cur > a) ret = min(ret, pre);
      pre *= -1;
      a = f(i - 1, 0, 1);
      if (cur > pre && cur < a) ret = min(ret, a);
      a = f(i - 1, 1, 1);
      if (cur < pre && cur > a) ret = min(ret, pre);
    }
    return dp[i][flag][sgn] = ret;
  }
  void dfs(int i, int flag, int sgn) {
    if (sgn) arr[i] *= -1;
    if (i == 0) return;
    int cur = arr[i];
    int ret = f(i, flag, sgn);
    if (flag) {
      int pre = arr[i - 1];
      int a = f(i - 1, 0, 0);
      if (cur > pre && cur > a && ret == pre) {
        dfs(i - 1, 0, 0);
        return;
      }
      a = f(i - 1, 1, 0);
      if (cur > pre && cur > a && ret == a) {
        dfs(i - 1, 1, 0);
        return;
      }
      pre *= -1;
      a = f(i - 1, 0, 1);
      if (cur > pre && cur > a && ret == pre) {
        dfs(i - 1, 0, 1);
        return;
      }
      a = f(i - 1, 1, 1);
      if (cur > pre && cur > a && ret == a) {
        dfs(i - 1, 1, 1);
        return;
      }
    } else {
      int pre = arr[i - 1];
      int a = f(i - 1, 0, 0);
      if (cur > pre && cur < a && ret == a) {
        dfs(i - 1, 0, 0);
        return;
      }
      a = f(i - 1, 1, 0);
      if (cur < pre && cur > a && ret == pre) {
        dfs(i - 1, 1, 0);
      }
      pre *= -1;
      a = f(i - 1, 0, 1);
      if (cur > pre && cur < a && ret == a) {
        dfs(i - 1, 0, 1);
        return;
      }
      a = f(i - 1, 1, 1);
      if (cur < pre && cur > a && ret == pre) {
        dfs(i - 1, 1, 1);
        return;
      }
    }
  }
  void solve() {
    cas++;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> arr[i];
    if (f(n - 1, 0, 0) <= n)
      dfs(n - 1, 0, 0);
    else if (f(n - 1, 0, 1) <= n)
      dfs(n - 1, 0, 1);
    else if (f(n - 1, 1, 0) <= n)
      dfs(n - 1, 1, 0);
    else if (f(n - 1, 1, 1) <= n)
      dfs(n - 1, 1, 1);
    else {
      cout << "NO" << '\n';
      return;
    }
    cout << "YES" << '\n';
    for (int i = 0; i < n; i++) cout << arr[i] << ' ';
    cout << '\n';
  }
  int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  