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: Wrong Answer)

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

typedef long long ll;

int main() {
    int N;
    cin >> N;
    vector<int> v(N, 0), pos(N + 1, 0);
    for (int i = 0; i < N; i++) {
        cin >> v[i];
        pos[v[i]] = i;
    }

    set<int> s;
    ll ans = 0;
    for (int i = N; i > 0; i--) {
        auto itr = s.upper_bound(pos[i]);
        auto itl = itr;

        int r1 = itr == s.end() ? N : *(itr++);
        int r2 = itr == s.end() ? N : *itr;
        int l1 = itl == s.begin() ? -1 : *(--itl);
        int l2 = itl == s.begin() ? -1 : *(--itl);

        ans += ((l1 - l2) * (r1 - pos[i]) + (r2 - r1) * (pos[i] - l1)) * i;
        s.insert(pos[i]);
    }

    cout << ans << endl;
    return 0;
}