Problem:
Score : 700 points

Problem Statement

Snuke has a permutation (P_0,P_1,\cdots,P_{N-1}) of (0,1,\cdots,N-1).

Now, he will perform the following operation exactly once:

Choose K consecutive elements in P and sort them in ascending order.

Find the number of permutations that can be produced as P after the operation.

Constraints

2 \leq N \leq 200000

2 \leq K \leq N

0 \leq P_i \leq N-1

P_0,P_1,\cdots,P_{N-1} are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
P_0 P_1 \cdots P_{N-1}

Output

Print the number of permutations that can be produced as P after the operation.

Sample Input 1

5 3
0 2 1 4 3

Sample Output 1

2

Two permutations can be produced as P after the operation: (0,1,2,4,3) and (0,2,1,3,4).

Sample Input 2

4 4
0 1 2 3

Sample Output 2

1

Sample Input 3

10 4
2 0 1 3 7 5 4 6 8 9

Sample Output 3

6

Solution in Ruby (2.3.3) (judge verdict: Accepted)

class Heap
  attr_reader :size

  def initialize
    @size = 0
    @heap = []
  end

  def empty?
    @size.zero?
  end

  def push(val)
    index = size
    @size += 1

    while index > 0
      parent = (index - 1) / 2

      break if @heap[parent] <= val

      @heap[index] = @heap[parent]
      index = parent
    end

    @heap[index] = val
  end

  alias_method :<<, :push

  def pop
    val = @heap.first
    root = @heap[size - 1]
    @heap[size - 1] = nil
    @size -= 1
    index = 0

    while index * 2 + 1 < @size
      left = 2 * index + 1
      right = left + 1

      if right < size && @heap[right] < @heap[left]
        left = right
      end

      break if @heap[left] >= root

      @heap[index] = @heap[left]
      index = left
    end

    @heap[index] = root
    val
  end

  def top
    @heap.first
  end
end

N, K = gets.split.map(&:to_i)
P = gets.split.map(&:to_i)

list = []

sum = 0
ans = 0

min_heap = Heap.new
max_heap = Heap.new

no_sort = false
gone = Hash.new(false)
cur = -1
suc = 0

N.times do |i|
  if cur < P[i]
    suc += 1
  else
    suc = 1
  end

  cur = P[i]
  
  min_heap.push(P[i])
  max_heap.push(-P[i])

  next if K - 1 > i

  no_change = false

  if i >= K
    v = P[i - K]
    gone[v] = true

    while !max_heap.empty? && gone[-max_heap.top]
      max_heap.pop
    end

    if -max_heap.top == P[i] && min_heap.top == v
      no_change = true
    end

    while !min_heap.empty? && gone[min_heap.top]
      min_heap.pop
    end
  end

  if suc < K
    ans += 1 unless no_change
  else
    no_sort = true
  end
end

if no_sort
  ans += 1
end

puts ans