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: Accepted)

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

int main()
{
    long long N;
    cin >> N;

    vector<long long> P(N + 1, 0);
    for (long long i = 1; i <= N; i++)
    {
        long long p;
        cin >> p;
        P[p] = i;
    }

    set<long long> s;
    long long ans = 0;
    for (long long i = N; i > 0; i--)
    {
        s.insert(P[i]);
        if (s.size() < 2)
        {
            continue;
        }
        long long n1 = P[i] - 1;
        long long n2 = 0;
        long long m1 = N - P[i];
        long long m2 = 0;

        auto itr = s.find(P[i]);

        if (itr != s.begin())
        {
            itr--;
            n1 = P[i] - *itr - 1;
            n2 = *itr - 1;
            if (itr != s.begin())
            {
                n2 = *itr - *(--itr) - 1;
            }
        }

        itr = s.find(P[i]);
        if (++itr != s.end())
        {
            m1 = *itr - P[i] - 1;
            m2 = N - *itr;
            if (++itr != s.end())
            {
                m2 = *itr - *(--itr) - 1;
            }
        }

        long long count = 0;
        itr = s.find(P[i]);
        if (itr != s.begin())
        {
            count += (n2 + 1) * (m1 + 1);
        }
        if (++itr != s.end())
        {
            count += (m2 + 1) * (n1 + 1);
        }
        ans += i * count;
    }
    cout << ans << endl;
}