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;

int main(int argc, char const *argv[])
{
	int n;
	cin>>n;
	pair<int, int> p[n];
	for (int i = 0; i < n; ++i)
	{
		cin>>p[i].first;
		p[i].second=i;
	}
	sort(p, p+n);
	reverse(p, p+n);
	multiset<int> memo={-1,-1,n,n};
	long long ans=0LL;
	for (int i = 0; i < n; ++i)
	{
		auto itr=memo.insert(p[i].second);
		--itr;
		int l1=*itr;
		--itr;
		int l2=*itr;
		++itr,++itr,++itr;
		int r1=*itr;
		++itr;
		int r2=*itr;
		int cnt=(l1-l2)*(r1-p[i].second)+(p[i].second-l1)*(r2-r1);
		ans+=(long long)p[i].first*cnt;
	}
	cout<<ans<<endl;
	return 0;
}