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.

Examples

Input

5 3
0 2 1 4 3


Output

2


Input

4 4
0 1 2 3


Output

1


Input

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


Output

6
from heapq import heappush ,heappop
n,k=map(int,input().split())
p=list(map(int,input().split()))
cnt=1
flag=[1]*n
for i in range(1,n):
    if p[i]>p[i-1]:
        cnt+=1
        flag[i]=cnt
    else:
        cnt=1
h=flag.count(k)
if h!=0:
    h-=1
lst_m=[]
lst_M=[]
for i in range(k):
    heappush(lst_m,p[i])
    heappush(lst_M,-p[i])
data_m=[]
data_M=[]
count=0
for i in range(n-k):
    m=lst_m[0]
    M=lst_M[0]
    heappush(lst_m,p[k+i])
    heappush(lst_M,-p[k+i])
    heappush(data_m,p[i])
    heappush(data_M,-p[i])
    if p[i]==lst_m[0]:
        while data_m[0]==lst_m[0]:
            heappop(data_m)
            heappop(lst_m)
            if not data_m:
                break
        if -p[k+i]==lst_M[0]:
            count+=1
    if -p[i]==lst_M[0]:
        while data_M[0]==lst_M[0]:
            heappop(data_M)
            heappop(lst_M)
            if not data_M:
                break
print(n-k+1-count-h)