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 N = 1000 * 1000 + 13;
  int n;
  int p[N], a[N];
  int dp[N][2], pr[N][2];
  void solve() {
    scanf("%d", &n);
    for (int i = 0; i < int(n); ++i) scanf("%d", &p[i]);
    for (int i = 0; i < int(n); ++i)
      for (int j = 0; j < int(2); ++j) dp[i][j] = INF;
    dp[0][0] = dp[0][1] = -INF;
    for (int i = 0; i < int(n - 1); ++i)
      for (int j = 0; j < int(2); ++j)
        if (dp[i][j] != INF) {
          int x = j ? -p[i] : p[i];
          int y = dp[i][j];
          if (x < y) swap(x, y);
          for (int nj = 0; nj < int(2); ++nj) {
            int z = nj ? -p[i + 1] : p[i + 1];
            if (z > x) {
              if (dp[i + 1][nj] > y) {
                dp[i + 1][nj] = y;
                pr[i + 1][nj] = j;
              }
            } else if (z > y) {
              if (dp[i + 1][nj] > x) {
                dp[i + 1][nj] = x;
                pr[i + 1][nj] = j;
              }
            }
          }
        }
    int j = -1;
    for (int i = 0; i < int(2); ++i)
      if (dp[n - 1][i] != INF) j = i;
    if (j == -1) {
      puts("NO");
      return;
    }
    for (int i = n - 1; i >= 0; i--) {
      a[i] = j ? -p[i] : p[i];
      j = pr[i][j];
    }
    puts("YES");
    for (int i = 0; i < int(n); ++i) printf("%d ", a[i]);
    puts("");
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) solve();
  }
  