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;

#define i64 long long int
#define ran 101101

int n;
pair<int, int> a[ran];
set<int> st;
set<int> ::iterator lit, rit;

int main() {
	scanf("%d", &n);
	for(int i=1;i<=n;i++) {
		scanf("%d", &a[i].first);
		a[i].second = i;
	}
	sort(a+1, a+n+1);
	
	i64 ans = 0;
	st.insert(0);
	st.insert(n+1);
	for(int i=n;i>0;i--) {
		int val = a[i].first;
		int pos = a[i].second;
		
		rit = st.lower_bound(pos);
		lit = rit;
		lit--;
		int lpos = (*lit);
		int rpos = (*rit);
		if (lpos>0) {
			lit--;
			ans += 1LL * (lpos - (*lit)) * (rpos - pos) * val;
		}
		if (rpos<=n) {
			rit++;
			ans += 1LL * ((*rit) - rpos) * (pos - lpos) * val;
		}
		
		st.insert(pos);
	}
	cout<<ans<<endl;
	
	return 0;
}