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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

N=int(input())
P=list(map(int,input().split()))
total_ans=0
for i in range(N):
    ans=0
    a=b=c=d=-1
    j=i-1
    while j>=0 and a==-1:
        if P[j]>P[i]:
            if b==-1:
                b=j
            elif a==-1:
                a=j
        j-=1
    j=i+1
    while j<=N-1 and d==-1:
        if P[j]>P[i]:
            if c==-1:
                c=j
            elif d==-1:
                d=j
        j+=1

#    print(f'a={a},b={b},c={c},d={d}')

    if a!=-1!=b:
        if c!=-1!=d:
            ans+=(b-a)*(c-i)+(i-b)*(d-c)
        elif c!=-1==d:
            ans+=(b-a)*(c-i)+(i-b)*(N-c)
        elif c==-1==d:
            ans+=(b-a)*(N-i)
    elif a==-1!=b:
        if c!=-1!=d:
            ans+=(b+1)*(c-i)+(i-b)*(d-c)
        elif c!=-1==d:
            ans+=(b+1)*(c-i)+(i-b)*(N-c)
        elif c==-1==d:
            ans+=(b+1)*(N-i)
    elif a==-1==b:
        if c!=-1!=d:
            ans+=(i+1)*(d-c)
        elif c!=-1==d:
            ans+=(i+1)*(N-c)
    
#    print(ans)
    total_ans+=P[i]*ans
print(total_ans)