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;
mt19937 rng(std::chrono::system_clock::now().time_since_epoch().count());
const int N = 2e5 + 1;
int n,k,p[N],bit[N],ans;
void add(int at,int val){
	for(int cur = at;cur < N;cur+=cur&(-cur))bit[cur]+=val;
}
int get(int at){
	int ret = 0;
	for(int cur = at;cur;cur-=cur&(-cur))ret+=bit[cur];
	return ret;
}
bool did;
int main(){
	//freopen("input.txt","r",stdin);
	scanf("%d%d",&n,&k);
	for(int i = 0;i < n;i++)scanf("%d",p + i),p[i]++;
	bool lastf = 0;
	long long inv = 0;
	for(int i = 0;i < k;i++){
		inv+=-get(p[i]) + i;
		add(p[i],1);
	}
	if(get(p[0]) - 1)lastf = 1;
	ans+=!!inv;
	did|=!inv;
	for(int i = k;i < n;i++){
		add(p[i - k],-1);
		inv-=get(p[i - k]);
		inv+=k - 1 - get(p[i]);
		add(p[i],1);
		bool curf = get(p[i - k + 1]) - 1;
		bool cure = k - get(p[i]);
		if(lastf || cure)ans+=!!inv;
		lastf = curf;
		did|=!inv;
	}
	printf("%d\n",ans + did);
}