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
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 233;
set<int> s;
int a[maxn];
bool ok[maxn];
int main()
{
	int n,cnt=0,k;
	cin>>n>>k;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int i=2;i<=k;i++)
		if(a[i]>a[i-1])cnt++;
	for(int i=1;i<=n-k+1;i++){
		if(cnt==k-1)ok[i] = 1;
		if(a[i]<a[i+1])cnt--;
		if(a[i+k]>a[i+k-1])cnt++;
	}
//	for(int i=1;i<=n-k+1;i++)cout<<ok[i]<<" ";cout<<endl;
	int ans = 1,first = ok[1];
	for(int i=2;i<=k;i++)s.insert(a[i]);
	for(int i=1;i<n-k+1;i++){
		int l = (*s.begin()),r = *s.rbegin();
//		cout<<i<<" "<<l<<" " << r << endl;
		if(a[i]<l&&r<a[i+k]){
			
		}
		else{
			if(ok[i+1]){
				if(!first){
					first = 1;
					ans++;
				}
			}
			else{
				ans++;
			}
		}
		s.erase(a[i+1]);
		s.insert(a[i+k]);
	}
	cout << ans << endl; 
	return 0;
} 