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 N = 2e6 + 5;
  int n, a[N], p[N], f[N];
  void solve() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 1; i < 2 * n; i++) p[i] = -1, f[i] = n + 1;
    f[0] = f[1] = -n;
    for (int i = 1; i <= n - 1; i++)
      for (int x = 0; x <= 1; x++)
        for (int y = 0; y <= 1; y++) {
          int v[2] = {f[i * 2 - 2 + y], a[i - 1] * (2 * y - 1)};
          for (int j = 0; j <= 1; j++)
            if (v[j] <= a[i] * (2 * x - 1) && v[!j] < f[i * 2 + x])
              f[i * 2 + x] = v[!j], p[i * 2 + x] = i * 2 - 2 + y;
        }
    if (f[2 * n - 1] <= n) {
      puts("YES");
      for (int i = n - 1, x = 2 * n - 1; ~i; i--, x = p[x])
        if (!(x & 1)) a[i] *= -1;
      for (int i = 0; i < n; i++) printf("%d ", a[i]);
      puts("");
    } else {
      puts("NO");
    }
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  