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)

class PriorityQueue
  def initialize(array = [])
    @data = []
    array.each{|a| add(a)}
  end
  
  def add(element)
    @data.push(element)
    bottom_up
  end
  
  # @return Integer
  def peek
    @data[0]
  end

  # @return Integer
  def poll
    if size == 0
      return nil
    elsif size == 1
      return @data.pop
    else
      max = @data[0]
      @data[0] = @data.pop
      top_down
      return max
    end
  end
  
  # @return Integer
  def size
    @data.size
  end

  private

  def swap(i, j)
    @data[i], @data[j] = @data[j], @data[i]
  end

  
  def bottom_up
    idx = size - 1
    while(has_parent?(idx))
      parent_idx = parent_idx(idx)
      if @data[parent_idx] < @data[idx]
        swap(parent_idx, idx)
        idx = parent_idx
      else
        return
      end
    end
  end
  
  def top_down
    idx = 0
    while (has_child?(idx))
      a = left_child_idx(idx)
      b = right_child_idx(idx)
      if @data[b].nil?
        c = a
      else
        c = @data[a] >= @data[b] ? a : b
      end
      if @data[idx] < @data[c]
        swap(idx, c)
        idx = c
      else
        return
      end
    end
  end
  

  # @param Integer
  # @return Integer
  def parent_idx(idx)
    (idx - (idx.even? ? 2 : 1)) / 2
  end

  # @param Integer
  # @return Integer
  def left_child_idx(idx)
    (idx * 2) + 1
  end

  # @param Integer
  # @return Integer
  def right_child_idx(idx)
    (idx * 2) + 2
  end

  # @param Integer
  # @return Boolent
  def has_child?(idx)
    ((idx * 2) + 1) < @data.size
  end

  def has_parent?(idx)
    idx > 0
  end
end

def sl(i,j)
  return m[i][j] unless m[i][j].nil?
  
end

n=gets.to_i
a=gets.split.map(&:to_i).unshift(0)
m=Array.new(n+1){Array.new()}
s=0
1.upto(n-1){|i|
  pq=PriorityQueue.new()
  pq.add(a[i])
  (i+1).upto(n){|j| 
    pq.add(a[j])
    y=0
    if m[i][j].nil?
      x=pq.poll
      y=pq.peek
      pq.add(x)
      m[i][j]=y
    else
      y=m[i][j]
    end
    s+=y 
  }}
p s