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 Ruby (2.3.3) (judge verdict: Accepted)

n = gets.to_i
p = gets.split.map &:to_i

idx = []

p.each_with_index{ |x, i| idx[x-1] = i }

keys = [idx[-1]]
count = 0

(idx.size - 2).downto(0){ |i|
	t_idx = keys.bsearch_index{ |x| x > idx[i] }
	t_idx = keys.size if t_idx == nil

	# 左の1個を含めた区間
	if t_idx >= 1 then
		l = 1
		r = 1

		if t_idx >= 2 then
			l = keys[t_idx - 1] - keys[t_idx - 2]
		else
			l = keys[t_idx - 1] + 1
		end
		
		if keys.size - 1 >= t_idx then
			r = keys[t_idx] - idx[i]
		else
			r = p.size - idx[i]
		end

		count += l * r * (i + 1)
	end

	# 右の1個を含めた区間
	if keys.size - 1 >= t_idx then
		l = 1
		r = 1

		if keys.size - 2 >= t_idx then
			r = keys[t_idx + 1] - keys[t_idx]
		else
			r = p.size - keys[t_idx]
		end
		
		if t_idx >= 1 then
			l = idx[i] - keys[t_idx - 1]
		else
			l = idx[i] + 1
		end

		count += l * r * (i + 1)
	end
	
	keys.insert(t_idx, idx[i])
}

puts count