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<iostream>
#include<set>
using namespace std;
int N,P[1<<17],A[1<<17];
set<int>S;
main()
{
	cin>>N;
	for(int i=0;i<N;i++)
	{
		cin>>P[i];
		A[P[i]]=i;
	}
	S.insert(A[N]);
	long ans=0;
	for(int i=N-1;i;i--)
	{
		S.insert(A[i]);
		set<int>::iterator id=S.find(A[i]);
		set<int>::iterator L=id,R=id;
		long B,AA,a,b;
		R++;
		if(R==S.end())
		{
			a=N-A[i];
			b=0;
		}
		else
		{
			int t=*R;
			a=t-A[i];
			R++;
			if(R==S.end())b=N-t;
			else b=*R-t;
		}
		if(L==S.begin())
		{
			AA=A[i]+1;
			B=0;
		}
		else
		{
			L--;
			int t=*L;
			AA=A[i]-t;
			if(L==S.begin())B=t+1;
			else
			{
				L--;
				B=t-*L;
			}
		}
		ans+=(AA*b+a*B)*i;
	}
	cout<<ans<<endl;
}