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;
  using namespace std;
  const long long N = 0 ? 15 : 1000005;
  int n, a[N];
  int dp[N][2][2];
  pair<int, int> pre[N][2][2];
  int main(int argc, const char *argv[]) {
    int t;
    scanf("%d", &t);
    while (t--) {
      scanf("%d", &n);
      for (int(i) = (1); (i) <= (long long)(n); (i)++) scanf("%d", &a[i]);
      for (int(i) = (0); (i) <= (long long)(1); (i)++)
        for (int(j) = (0); (j) <= (long long)(1); (j)++) dp[0][i][j] = -1e9;
      a[0] = 1e9;
      for (int(i) = (1); (i) <= (long long)(n); (i)++) {
        for (int(pos) = (0); (pos) <= (long long)(1); (pos)++) {
          for (int(sign) = (0); (sign) <= (long long)(1); (sign)++) {
            int &ans = dp[i][pos][sign];
            ans = 1e9;
            int z = sign ? a[i] : -a[i];
            for (int(psig) = (0); (psig) <= (long long)(1); (psig)++)
              for (int(ppos) = (0); (ppos) <= (long long)(1); (ppos)++) {
                int x = psig ? a[i - 1] : -a[i - 1];
                int y = dp[i - 1][ppos][psig];
                if (ppos) {
                  swap(x, y);
                }
                if (z > x && !pos) {
                  if (ans > y) {
                    ans = y;
                    pre[i][pos][sign] = {ppos, psig};
                  }
                } else if (z > y && pos) {
                  if (ans > x) {
                    ans = x;
                    pre[i][pos][sign] = {ppos, psig};
                  }
                }
              }
          }
        }
      }
      int pos = -1, sign = -1;
      for (int(p) = (0); (p) <= (long long)(1); (p)++)
        for (int(second) = (0); (second) <= (long long)(1); (second)++)
          if (dp[n][p][second] != 1e9) {
            pos = p;
            sign = second;
          }
      if (pos == -1) {
        printf("NO\n");
        continue;
      }
      for (long long(i) = (n); (i) >= (1); --(i)) {
        a[i] = sign ? a[i] : -a[i];
        pair<int, int> nxt = pre[i][pos][sign];
        pos = nxt.first;
        sign = nxt.second;
      }
      printf("YES\n");
      for (int(i) = (1); (i) <= (long long)(n); (i)++) printf("%d ", a[i]);
      printf("\n");
    }
    return 0;
  }
  