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;

#define rep(i, N) for (int i = 0; i < N; i++)

using ll = long long int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int main()
{
    int N;
    cin >> N;
    vector<int> P(N);
    rep(i, N)
    {
        cin >> P[i];
        P[i]--;
    }

    vector<int> P2i(N);
    rep(i, N)
    {
        P2i[P[i]] = i;
    }

    vector<bool> is_checked(N, false);
    is_checked[P2i[N - 1]] = true;
    set<int> s;
    s.insert(P2i[N - 1]);
    ll sum = 0;
    for (int p = N - 2; p >= 0; p--)
    {
        int i_p = P2i[p];
        int i_ll, i_l, i_r, i_rr;

        auto it = s.insert(i_p).first;
        auto it_first = s.begin();
        auto it_last = s.end();
        it_last--;
        if (it != it_first)
        {
            auto it_l = it;
            it_l--;
            i_l = *it_l;
            if (it_l != it_first)
            {
                it_l--;
                i_ll = *it_l;
            }
            else
                i_ll = -1;
        }
        else
            i_l = i_ll = -1;

        if (it != it_last)
        {
            auto it_r = it;
            it_r++;
            i_r = *it_r;
            if (it_r != it_last)
            {
                it_r++;
                i_rr = *it_r;
            }
            else
                i_rr = N;
        }
        else
            i_r = i_rr = N;

        ll dll = i_l - i_ll;
        ll dl = i_p - i_l;
        ll dr = i_r - i_p;
        ll drr = i_rr - i_r;
        sum += (dll * dr + dl * drr) * (p + 1);
    }
    cout << sum << endl;
}