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 C (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#include <math.h>


int main(void)
{
	long int i, j;
	long int n, k;
	long int sum;
	long int max;
	long int second;
	long int count;
	long int p[100001];
	long int b[100];
	long int c[100];
	
	scanf("%ld", &n);
	
	for(i = 1; i <= n; i++) {
		scanf("%ld", &p[i]);
	}
	
	sum = 0;
	
	for(i = 1; i <= n - 1; i++) {
		if(p[i] > p[i + 1]) {
			max = p[i];
			second = p[i + 1];
		} else {
			max = p[i + 1];
			second = p[i];
		}
		sum += second;
		for(j = i + 2; j <= n; j++) {
			if(max < p[j]) {
				second = max;
				max = p[j];
			} else if(second < p[j]) {
				second = p[j];
			}
			sum += second;
		}
		
	}
	
	printf("%ld\n", sum);
	
	
	return 0;
}