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<iostream>
#include<cstdio>
const int N=2e5+5;
int L[N],R[N];
int stk[N],top,n;
int a[N],k,f[N];
int main() {
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++) {
        scanf("%d",a+i);
        if(i==1||a[i]<a[i-1]) f[i]=1;
        else f[i]=1+f[i-1];
        while(top&&a[stk[top-1]]<=a[i]) --top;
        if(!top) L[i]=1;
        else L[i]=stk[top-1]+1;
        stk[top++]=i;
    }
    top=0;
    for(int i=n;i;i--) {
        while(top&&a[stk[top-1]]>=a[i]) --top;
        if(!top) R[i]=n;
        else R[i]=stk[top-1]-1;
        stk[top++]=i;
    }
    int ans=0;
    int cnt=0;
    for(int i=k;i<=n;i++) {
        if(f[i]>=k) {
            cnt=1;
            continue;
        }
        if(i>k&&R[i-k]>=i&&L[i]<=i-k) continue;
        ans++;
    }
    printf("%d\n",ans+cnt);
    return 0;
}
