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"

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define REP(i, a, b) for (int i = (a); i < (b); ++i)

using namespace std;
using ll = long long;

void Main() {
  int n;
  cin >> n;
  vector<int> p(n);
  rep(i, n) cin >> p[i];
  vector<int> idx(n);
  rep(i, n) {
    p[i]--;
    idx[p[i]] = i;
  }

  set<int> s;
  ll ans = 0;
  for (int x = n - 1; x >= 0; --x) {
    int i = idx[x];
    s.insert(i);
    vector<int> l(2, -1);
    {
      auto it = s.find(i);
      rep(j, 2) {
        if (it == s.begin()) break;
        --it;
        l[j] = *it;
      }
    }
    vector<int> r(2, n);
    {
      auto it = s.find(i);
      rep(j, 2) {
        ++it;
        if (it == s.end()) break;
        r[j] = *it;
      }
    }
    ans += (ll)(i - l[0]) * (r[1] - r[0]) * (x + 1);
    ans += (ll)(r[0] - i) * (l[0] - l[1]) * (x + 1);
  }
  cout << ans << endl;
}

int main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(15);
  Main();
  return 0;
}