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 <set>
#include <iostream>
#include <vector>

int main() {
    int n;
    std::cin >> n;
    std::vector<int> p_to_i(n+1);
    for (int i = 0; i < n; ++i) {
        int p;
        std::cin >> p;
        p_to_i[p] = i;
    }
    std::set<int> indices;
    for (int i = 0; i < n; ++i) {
        indices.insert(i);
    }
    long long ans = 0;
    for (int p = 1; p < n; ++p) {
        int i = p_to_i[p];
        const auto it = indices.find(i);
        auto next = std::next(it);
        if (it == indices.begin()) {
            int r1 = *next;
            int r2 = (std::next(next) == indices.end()) ? n : *std::next(next);
            ans += p * (r2-r1) * (i+1);
        } else if (std::next(it) == indices.end()) {
            auto prev = std::prev(it);
            int l1 = *prev;
            int l2 = (prev == indices.begin()) ? -1 : *std::prev(prev);
            ans += p * (l1-l2) * (n-i);
        } else {
            auto next = std::next(it);
            auto prev = std::prev(it);
            auto l1 = *prev;
            auto r1 = *next;
            int l2 = (prev == indices.begin()) ? -1 : *std::prev(prev);
            int r2 = (std::next(next) == indices.end()) ? n : *std::next(next);
            ans += p * ((l1-l2) * (r1-i) + (i-l1) * (r2-r1));
        }

        indices.erase(i);
    }
    std::cout << ans << std::endl;
}