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 T, n, p[1000005], dp[1000005][2], g[1000005][2], tans[1000005];
  void upd(int &ans, int &ans2, int x, int u, int v, int fl) {
    if (u > v) swap(u, v);
    if (x >= v) {
      if (u < ans) {
        ans = u;
        ans2 = fl;
      }
    } else if (x >= u) {
      if (v < ans) {
        ans = v;
        ans2 = fl;
      }
    }
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
      dp[1][0] = dp[1][1] = -1e9;
      for (int i = 2; i <= n; i++) {
        dp[i][0] = dp[i][1] = 1e9;
        upd(dp[i][0], g[i][0], -p[i], -p[i - 1], dp[i - 1][0], 0);
        upd(dp[i][0], g[i][0], -p[i], p[i - 1], dp[i - 1][1], 1);
        upd(dp[i][1], g[i][1], p[i], -p[i - 1], dp[i - 1][0], 0);
        upd(dp[i][1], g[i][1], p[i], p[i - 1], dp[i - 1][1], 1);
      }
      if (dp[n][1] > 1e8) {
        printf("NO\n");
        continue;
      }
      printf("YES\n");
      int fl = 1;
      for (int i = n; i > 0; i--) {
        tans[i] = p[i] * (fl * 2 - 1);
        fl = g[i][fl];
      }
      for (int i = 1; i <= n; i++) printf("%d ", tans[i]);
      printf("\n");
    }
    return 0;
  }
  