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;

typedef long long int ll;
ll MOD = 1000000007;
ll INFL = 1ll << 60;
ll INF = 1 << 28;

// ====================================================================

int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) { cin >> v[i]; }

  vector<int> pos(n + 1);  // それぞれの数字がどこにあるか？
  for (int i = 0; i < n; i++) pos[v[i]] = i;

  ll ans = 0;
  multiset<int> st = {-1, -1, n, n};  // 自分より大きいやつの位置
  for (ll i = n; i > 0; i--) {
    st.insert(pos[i]);

    // for (auto x : st)
    //   cout << x << " ";
    // cout << endl;

    auto now = st.lower_bound(pos[i]);
    // 左側を含むやつ
    // cout << "prev " << *now << " " << *prev(now, 1) << " " << *prev(now, 2) << endl;
    ans += i * (*prev(now, 1) - *prev(now, 2)) * (*next(now, 1) - *now);
    // 右側を含むやつ
    ans += i * (*now - *prev(now, 1)) * (*next(now, 2) - *next(now, 1));
    // cout << i << " " << ans << endl;
  }
  cout << ans << endl;
}