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.Arrays;
import java.util.Scanner;
import java.util.TreeSet;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int P[] = new int[N];
		int A[] = new int[N + 1];
		for(int i = 0 ; i < N ; ++i) {
			P[i] = sc.nextInt();
			A[P[i]] = i + 1;
		}
		TreeSet<Integer> phigh = new TreeSet<>();
		phigh.add(0);
		phigh.add(N + 1);
		phigh.add(A[N]);
		long S = 0;
		//X [i,j]  = P => [i,j] P未満かつP以上が一つ
		for(int n = N - 1; n >= 1 ; --n) {
			int Xi = A[n];
			long y = phigh.higher(Xi);
			long z = N + 1;
			if(y < N + 1) {
				z = phigh.higher((int)y);				
			}
			long x = phigh.lower(Xi);
			long w = 0;
			if(x != 0) {
				w = phigh.lower((int)x);				
			}
			long C = (x - w) * (y - Xi) + (Xi - x) * (z - y);
			S += C * n;
			phigh.add(Xi);
		}
		System.out.println(S);
	}
}