Problem:
Score: 500 points

Problem Statement

Given is a permutation P of \{1, 2, \ldots, N\}.

For a pair (L, R) (1 \le L \lt R \le N), let X_{L, R} be the second largest value among P_L, P_{L+1}, \ldots, P_R.

Find \displaystyle \sum_{L=1}^{N-1} \sum_{R=L+1}^{N} X_{L,R}.

Constraints

2 \le N \le 10^5

1 \le P_i \le N

P_i \neq P_j (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 P_2 \ldots P_N

Output

Print \displaystyle \sum_{L=1}^{N-1} \sum_{R=L+1}^{N} X_{L,R}.

Sample Input 1

3
2 3 1

Sample Output 1

5

X_{1, 2} = 2, X_{1, 3} = 2, and X_{2, 3} = 1, so the sum is 2 + 2 + 1 = 5.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

30

Sample Input 3

8
8 2 7 3 4 5 6 1

Sample Output 3

136

Solution in Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.math.*;
import java.util.*;
import java.util.function.*;
import java.util.stream.*;

public class Main {
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    List<Long> p = new ArrayList<>();
    for (int i = 0; i < n; i++) {
      p.add(sc.nextLong());
    }
    long result = 0;
    for (int l = 0, len = n - 1; l < len; l++) {
      long pl1 = p.get(l);
      long pl2 = p.get(l + 1);
      long max = Math.max(pl1, pl2);
      long max2 = Math.min(pl1, pl2);
      result += max2;
      for (int r = l + 2; r < n; r++) {
        long pr = p.get(r);
        if (pr > max) {
          max2 = max;
          max = pr;
        } else if (pr > max2) {
          max2 = pr;
        }
        result += max2;
      }
    }
    System.out.println(result);
  }
}