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())
A = list(map(int,input().split()))

idx = [0] * (N+3)
cnt = 1
for a in A:
    idx[a] = cnt
    cnt += 1
idx[N+1] = N+1
idx[N+2] = 0

sets = [0,N+1]

ans = 0
import bisect
for i in reversed(range(1,N+1)):
    index = idx[i]
    tmplist = sets
    L = len(tmplist)
    left1 = max(bisect.bisect_left(tmplist,index)-1, 0)
    left2 = max(left1-1,0)
    right1 = min(bisect.bisect_left(tmplist,index), L-1)
    right2 = min(right1+1, L-1)
    
    l1 = tmplist[left1]
    l2 = tmplist[left2]
    r1 = tmplist[right1]
    r2 = tmplist[right2]
    ans += i * ((r1-index)*(l1-l2) + (index-l1)*(r2-r1))
    insert_index = bisect.bisect_left(tmplist,index)
    sets = tmplist[:insert_index] + [index] + tmplist[insert_index:]
    
    
print(ans)