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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

n=int(input())
p=list(map(int,input().split()))

def find_second(p,s,e):
    max1=0
    pos1=0
    for i in range(s-1,e):
        if p[i]>max1:
            max1=p[i]
            pos1=i
    #print(pos1,max1)
    max2=0
    pos2=0
    for i in range(s-1,e):
        if p[i]>max2 and i!=pos1:
            max2=p[i]
            pos2=i
    #print(pos2,max2)
    return p[pos2]

ans=0
for j in range(1,n):
    for k in range(j+1,n+1):
        ans+=find_second(p,j,k)
print(ans)