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: Time Limit Exceeded)

n=gets.to_i
a=gets.chomp.split(" ").map(&:to_i)
b=[]
re=0
(n-1).times do |i|
  if a[i]<a[i+1]
    b<<a[i]
    b<<a[i+1]
  else
    b<<a[i+1]
    b<<a[i]
  end
  re+=b[0]
  (n-i-2).times do |k|
    if a[i+k+2] <= b[0]
      re+=b[0]
      next
    elsif a[i+k+2] > b[0] && a[i+k+2] <=b[1]
      b.shift
      b.unshift(a[i+k+2])
      re+=b[0]
    else
      b.shift
      b << a[i+k+2]
      re+=b[0]
    end
  end
  b=[]
end
  puts re