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.*;

public class Main {
	Scanner sc;
	int N;
	int[][] P;
	
	public Main() {
		sc = new Scanner(System.in);
	}
	
	private void calc() {
		N = sc.nextInt();
		P = new int[N][2];
		for (int i = 0; i < N; i++) {
			P[i][0] = i;
			P[i][1] = sc.nextInt();
		}
		Arrays.sort(P, (a,b) -> -Integer.compare(a[1],b[1]));
		TreeSet<Integer> t = new TreeSet<>();
		long s = 0;
		for (int i = 0; i < N; i++) {
			int ind = P[i][0];
			t.add(ind);
			Iterator<Integer> tail = t.tailSet(ind, false).iterator();
			int c = tail.hasNext()?tail.next():N;
			int d = tail.hasNext()?tail.next():N;
			Iterator<Integer> head = t.headSet(ind, false).descendingIterator();
			int b = head.hasNext()?head.next():-1;
			int a = head.hasNext()?head.next():-1;
			s += ((long)P[i][1])* ( (c-ind)*(b-a) + (d-c)*(ind-b) );
		}
		System.out.println(s);
	}
	
	public static void main(String[] args) {
		new Main().calc();
	}
}