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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MIN(a, b) (a<b?a:b)
#define MAX(a, b) (a>b?a:b)


int main(int argc, char const* argv[])
{
	int n, i, j;
	int p[100000];
	long long ans=0;
	int fst, scnd;
	scanf("%d", &n);
	for(i=0; i<n; i++) {
		scanf("%d", p+i);
	}
	for(i=0; i<n-1; i++) {
		fst = MAX(p[i], p[i+1]);
		scnd = MIN(p[i], p[i+1]);
		ans += (long long)scnd;
		for(j=i+2; j<n; j++) {
			if(p[j] > fst) {
				scnd = fst;
				fst = p[j];
			} else if(p[j] > scnd) {
				scnd = p[j];
			}
			ans += (long long)scnd;
		}
	}
	printf("%lld\n", ans);
	
	return 0;
}