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;
  long long a[1000005], dp[1000005][2];
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long T;
    cin >> T;
    while (T--) {
      long long n;
      cin >> n;
      for (long long i = 1; i <= n; i++) cin >> a[i];
      a[0] = 1e9, dp[0][0] = -1e9;
      for (long long i = 1; i <= n; i++) dp[i][0] = dp[i][1] = 1e9;
      for (long long i = 1; i <= n; i++) {
        long long x = -a[i - 1], y = dp[i - 1][0];
        if (a[i] > x) dp[i][1] = min(dp[i][1], y);
        if (a[i] > y) dp[i][1] = min(dp[i][1], x);
        if (-a[i] > x) dp[i][0] = min(dp[i][0], y);
        if (-a[i] > y) dp[i][0] = min(dp[i][0], x);
        x = a[i - 1], y = dp[i - 1][1];
        if (a[i] > x) dp[i][1] = min(dp[i][1], y);
        if (a[i] > y) dp[i][1] = min(dp[i][1], x);
        if (-a[i] > x) dp[i][0] = min(dp[i][0], y);
        if (-a[i] > y) dp[i][0] = min(dp[i][0], x);
      }
      if (dp[n][1] <= n || dp[n][0] <= n) {
        cout << "YES\n";
        long long now = 0;
        if (dp[n][1] <= n) now = 1;
        for (long long i = n; i >= 1; i--) {
          if (now == 1) {
            long long x = -a[i - 1], y = dp[i - 1][0];
            if (a[i] > x && dp[i][1] == y) now = 0;
            if (a[i] > y && dp[i][1] == x) now = 0;
            x = a[i - 1], y = dp[i - 1][1];
            if (a[i] > x && dp[i][1] == y) now = 1;
            if (a[i] > y && dp[i][1] == x) now = 1;
          } else {
            long long x = -a[i - 1], y = dp[i - 1][0];
            if (-a[i] > x && dp[i][0] == y) now = 0;
            if (-a[i] > y && dp[i][0] == x) now = 0;
            x = a[i - 1], y = dp[i - 1][1];
            if (-a[i] > x && dp[i][0] == y) now = 1;
            if (-a[i] > y && dp[i][0] == x) now = 1;
            a[i] = -a[i];
          }
        }
        for (long long i = 1; i <= n; i++) cout << a[i] << " ";
        cout << "\n";
      } else
        cout << "NO\n";
    }
    return 0;
  }
  