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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

/**
 * Second Sum
 */
public class Main {

    public static void main(String[] args) throws Exception {
        try (Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();
            int[] P = new int[N];
            for (int i = 0; i < N; i++) {
                P[i] = sc.nextInt();
            }

            int[] Q = new int[N];
            for (int i = 0; i < N; i++) {
                Q[P[i] - 1] = i;
            }

            long sum = 0;

            List<Integer> index = new ArrayList<>();
            index.add(Q[N - 1]);

            for (int i = N - 2; i >= 0; i--) {
                int p = Q[i];

                int c = -(Collections.binarySearch(index, p) + 1);

                int a1 = c - 1 >= 0 ? index.get(c - 1) : -1;
                int a2 = c - 2 >= 0 ? index.get(c - 2) : -1;
                int b1 = c < index.size() ? index.get(c) : N;
                int b2 = c + 1 < index.size() ? index.get(c + 1) : N;

                if (a1 != -1) {
                    sum += (long) (i + 1) * (a1 - a2) * (b1 - p);
                }

                if (b1 != N) {
                    sum += (long) (i + 1) * (p - a1) * (b2 - b1);
                }

                index.add(c, p);
            }

            System.out.println(sum);
        }
    }

}