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: Time Limit Exceeded)

n, k = gets.split.map(&:to_i)
p = gets.split.map(&:to_i)
inv = [0]
group = [0]
1.upto(n - 1) do |i|
    if p[i - 1] > p[i] then
        inv[i] = inv[i - 1] + 1
        group[i] = i
    else
        inv[i] = inv[i - 1]
        group[i] = group[i - 1]
    end
end

count = 0
exist_id = false
if inv[0] == inv[k - 1] then
    exist_id = true
else
    count = 1
end
1.upto(n - k) do |i|
    if inv[i] == inv[i + k - 1] then
        exist_id = true
    else
        if p[(i - 1)..(i + k - 2)].min != p[i - 1] || p[i..(i + k - 1)].max != p[i + k - 1] then
            count += 1
        end
    end
end

if exist_id then
    puts count + 1
else
    puts count
end