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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cassert>
#include <cstdint>
#include <iostream>
#include <iomanip>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <numeric>
#include <bitset>

using namespace std;

using ll =  long long;
using Pll = pair<ll, ll>;
using Pii = pair<int, int>;

constexpr ll MOD = 1000000007;
constexpr long double EPS = 1e-10;
constexpr int dyx[4][2] = {
    { 0, 1}, {-1, 0}, {0,-1}, {1, 0}
};

int main() {
    std::ios::sync_with_stdio(false); cin.tie(nullptr);
    int n;
    cin >> n;
    vector<Pii> a(n); // {Pi, idx}
    for(int i=0;i<n;++i) {
        cin >> a[i].first;
        a[i].second = i;
    }
    sort(a.rbegin(), a.rend());

    int p, idx;
    set<int> used, used2;
    ll ans = 0;
    vector<vector<int>> lowers(2, vector<int>(n+1, -1)), uppers(2, vector<int>(n+1, n));
    for(int i=0;i<n;++i) {
        tie(p, idx) = a[i];
        if(i == 0) {
            used.insert(idx);
            used2.insert(-idx);
            continue;
        }

        auto itr = used.lower_bound(idx);
        if(itr != used.end()) {
            uppers[0][p] = *itr;
            itr = used.upper_bound(*itr);
            if(itr != used.end()) {
                uppers[1][p] = *itr;
            }
        }
        itr = used2.lower_bound(-idx);
        if(itr != used2.end()) {
            lowers[0][p] = -*itr;
            itr = used2.upper_bound(*itr);
            if(itr != used2.end()) {
                lowers[1][p] = -*itr;
            }
        }
        cerr << p << ", " << idx << ": " << lowers[1][p] << " " << lowers[0][p] << " " << uppers[0][p] << " " << uppers[1][p] << endl;

        ans += (lowers[0][p]-lowers[1][p]) * (uppers[0][p] - idx) * p;
        ans += (uppers[1][p]-uppers[0][p]) * (idx - lowers[0][p]) * p;

        used.insert(idx);
        used2.insert(-idx);

    }
    cout << ans << endl;
}