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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin >> n;
    vector<int> p(n),top(n),second(n);
    unsigned long sum=0;

    cin >> p[0];

    for(int i=1; i<n; i++){
        cin >> p[i];
        if(p[i-1] <= p[i]){
            sum += p[i-1];
            top[i-1] = p[i];
            second[i-1] = p[i-1];
        }else{
            sum += p[i];
            second[i-1] = p[i];
            top[i-1] = p[i-1];
        }
        for(int j=2; j<=i; j++){
            if(p[i] >= top[i-j]){
                second[i-j] = top[i-j];
                top[i-j] = p[i];
            }else if(p[i] < top[i-j] && p[i] >= second[i-j]){
                second[i-j] = p[i];
            }
            //cout << top[i-j] << " " << second[i-j] << endl;
            sum += second[i-j];
        }
    }

    cout << sum << endl;
}