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 ll long long
using namespace std;
const int N=100005;
int n,a[N],pos[N];
int pre[N],nxt[N];
ll ans;
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]),pos[a[i]]=i;
	for (int i=0;i<=n+1;i++)
		pre[i]=max(0,i-1),nxt[i]=min(n+1,i+1);
	ll S=1ll*n*(n-1)/2;
   	for (int i=1;i<=n;i++){
   		ans+=S;
   		int x=pos[i];
   		S-=1ll*(pre[x]-pre[pre[x]])*(nxt[x]-x);
   		S-=1ll*(nxt[nxt[x]]-nxt[x])*(x-pre[x]);
   		pre[nxt[x]]=pre[x];
   		nxt[pre[x]]=nxt[x];
	}
	printf("%lld\n",ans);
}