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 = [int(i) for i in input().split()]
# P_dict = dict(zip([i for i in range(N)], P))

y = 0 # max
x = 0 # second max
counter = 0 # sum x
for l in range(N-1):
    for r in range(l+1, N):
        if l+1 == r: # init
            y = max(P[l], P[r])
            x = min(P[l], P[r])
        else:
            if P[r] > x and P[r] > y:
                x = y
                y = P[r]
            elif P[r] > x and P[r] < y:
                x = P[r]
        counter += x
        if x == N - 1:
            counter += x * (N - r - 1)
            break
print(counter)