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 N=2e5+10;
int a[N],f[N];
int n,k,sum,ans;
set<int>s;
int main(){
	scanf("%d%d",&n,&k);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	f[n]=1;
	for (int i=n-1;i>=1;i--) if (a[i]<a[i+1]) f[i]=f[i+1]+1; else f[i]=1;
	bool p=0;
	for (int i=1;i<=k;i++) s.insert(a[i]);
	for (int i=1;i<=n-k+1;i++) {
		if (f[i]>=k) {
			if (!p) ans++,p=1;
		} else if (i>1) {
			bool p=0;
			set<int>::iterator it;
			it=s.begin(); if (a[i-1]>(*it)) p=1;
			it=s.end(); it--;
			if (a[i+k-1]<(*it)) p=1;
			if (p) ans++;
		} else ans++;
		s.erase(a[i]); s.insert(a[i+k]);
	}
	printf("%d\n",ans);
}