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 second_choice(int *P,int N){
    int first=0;
    int second=0;
    int shelter;
    for(int i=0; i<N; i++){
        shelter = *(P+i);
        if(first < *(P+i)){
            shelter = first;
            first = *(P+i);
        }
        if(second < shelter){
            second = shelter;
        }
    }
    return second;
}

int main(void){
    int N;
    int P[100001];
    int i,k;
    int count=0;
    scanf("%d",&N);
    for(i=1; i<=N; i++){
        scanf("%d",&P[i]);
    }
    for(i=1; i<=N-1; i++){
        for(k=i+1; k<=N; k++){
            count += second_choice(&P[i],k-i+1);
        }
    }
    printf("%d",count);
}