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 <stdlib.h>
int compare(const void *a,const void *b){
    return *(int*)b-*(int*)a;
}
int main(){
    int n, x, y, ans;
    int p[10 * 10 * 10 * 10 * 10 + 999];
    int arr[3];
    scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%d",&p[i]);

    for(int l=1;l<=n-1;l++){
      x = p[l-1];
      y = -1;
      for(int r=l+1;r<=n;r++){
        arr[0] = x;
        arr[1] = y;
        arr[2] = p[r-1];
        qsort(arr,3,4,compare);
        y = arr[1];
        x = arr[0];
        ans += y;
      }
    }
    printf("%d\n",ans);
    return 0;
}