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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1e5;
int pre[N+2];
int post[N+2];
int p[N+2];
int loc[N+2];
int main()
{
    memset(pre,-1,sizeof pre);
    memset(post,-1,sizeof post);
    int n; cin >> n;
    for(int i = 1; i <= n; i++)
    {
        cin >> p[i];
        loc[p[i]] = i;
    }

    set<int> available;
    available.insert(0);
    available.insert(n+1);

    LL ans = 0;
    for(LL k = n; k >= 1; k--)
    {
        LL aft = *available.upper_bound(loc[k]);
        LL bef = *(--available.upper_bound(loc[k]));

        post[bef] = loc[k];
        pre[aft] = loc[k];
        pre[loc[k]] = bef;
        post[loc[k]] = aft;

        if(bef >= 1)
            ans += k*(bef-pre[bef])*(aft-loc[k]);
        if(aft <= n)
            ans += k*(loc[k]-bef)*(post[aft]-aft);
        
        available.insert(loc[k]);
    }
    cout<<ans<<endl;

    return 0;
}