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>

int secondbig(int a[], int size) {
    int max = 0;
    int secmax = 0;
    for(int i = 0; i < size; i++) {
        if(a[i] > max) max = a[i];
    }
    for(int i = 0; i < size; i++) {
        if(a[i] < max && a[i] > secmax) secmax = a[i];
    }
    return secmax;
}
    
int main() {
    int n;
    scanf("%d", &n);

    int p[n];
    for(int i = 0; i < n; i++) {
        scanf("%d", &p[i]);
    }
    int sum = 0;
    for(int l = 0; l < n-1; l++) {
        for(int r = l; r < n; r++) {
            int array[r-l+1];
            for(int i = 0; i < r-l+1; i++) {
                array[i] = p[l+i];
            }
            sum += secondbig(array, r-l+1);
        }
    }

    printf("%d\n", sum);

    return 0;
}