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 = 1000010;
  const int INf = 0x7f7f7f7f;
  int n, a[N], f[N][2], g[N][2];
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) {
        scanf("%d", a + i);
        f[i][0] = f[i][1] = INf;
      }
      f[1][0] = f[1][1] = -INf;
      for (int i = 2; i <= n; i++) {
        if (a[i] > a[i - 1] && f[i][0] > f[i - 1][0]) {
          f[i][0] = f[i - 1][0];
          g[i][0] = 0;
        }
        if (a[i] > f[i - 1][0] && f[i][0] > a[i - 1]) {
          f[i][0] = a[i - 1];
          g[i][0] = 0;
        }
        if (a[i] > -a[i - 1] && f[i][0] > f[i - 1][1]) {
          f[i][0] = f[i - 1][1];
          g[i][0] = 1;
        }
        if (a[i] > f[i - 1][1] && f[i][0] > -a[i - 1]) {
          f[i][0] = -a[i - 1];
          g[i][0] = 1;
        }
        if (-a[i] > a[i - 1] && f[i][1] > f[i - 1][0]) {
          f[i][1] = f[i - 1][0];
          g[i][1] = 0;
        }
        if (-a[i] > f[i - 1][0] && f[i][1] > a[i - 1]) {
          f[i][1] = a[i - 1];
          g[i][1] = 0;
        }
        if (-a[i] > -a[i - 1] && f[i][1] > f[i - 1][1]) {
          f[i][1] = f[i - 1][1];
          g[i][1] = 1;
        }
        if (-a[i] > f[i - 1][1] && f[i][1] > -a[i - 1]) {
          f[i][1] = -a[i - 1];
          g[i][1] = 1;
        }
      }
      if (f[n][0] == INf && f[n][1] == INf) {
        printf("NO\n");
      } else {
        printf("YES\n");
        int u = f[n][0] == INf ? 1 : 0;
        for (int i = n; i >= 1; i--) {
          a[i] = u ? -a[i] : a[i];
          u = g[i][u];
        }
        for (int i = 1; i <= n; i++) {
          printf("%d ", a[i]);
        }
        printf("\n");
      }
    }
    return 0;
  }
  