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: Wrong Answer)

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

max = n - k + 1

p_count = 0
del = 0

count = 1
t_p = p[0]
start_i = 0
a = []

(1...p.size).each{ |i|
	if p[i] > t_p then
		count += 1
	else
		if a.size == 0 || (a.size > 0 && a[-1][1] != i - 2) then
			a << [start_i, i - 1]
		end
		if count >= k then
			p_count += 1
			del += count - k
		end
		count = 1
		start_i = i
	end

	t_p = p[i]
}

a << [start_i, p.size - 1]
if count >= k then
	p_count += 1
	del += count - k
end

#print a, del, p_count, "\n"

a.each_cons(2){ |x|
	lx = x[0]
	rx = x[1]
	lx_len = lx[1] - lx[0] + 1
	rx_len = rx[1] - rx[0] + 1
	diff = rx[0] - lx[1] + 1
	
#	print [lx, rx, lx_len, rx_len, diff, del], "\n"
	
	if
		diff < k &&
		lx_len + diff + rx_len - 2 >= k &&
		(lx_len == 1 || (lx_len >= 2 && p[lx[1] - 1] < p[rx[0]])) &&
		(rx_len == 1 || (rx_len >= 2 && p[rx[0] + 1] > p[lx[1]]))
	then
#		print [k-diff, rx_len, lx_len], "\n"
		del += [k-diff, lx_len - 1, rx_len - 1, rx_len + lx_len + diff - 2 - k].min
	end
}

#print del

puts max - [p_count - 1, 0].max - del