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
/*
 * AtCoder AGC038 B Sorting a Segment
*/
#include <bits/stdc++.h>
using namespace std;
int n,k,a[200005],go[200005];
int main(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    int ans=0;
    go[0]=0;
    for(int i=2;i<=n;i++){
        if(a[i]>a[i-1])go[i]=go[i-1]+1;
        else go[i]=0;
        if(go[i]>=k-1)ans=1;
    }
    set<int> ss;
    for(int i=1;i<=k;i++)ss.insert(a[i]);
    if(go[k]!=k-1)ans++;
    for(int i=k+1;i<=n;i++){
        ss.insert(a[i]);
        if(*ss.begin()!=a[i-k]||*ss.rbegin()!=a[i]){
            if(go[i]<k-1){
                ans++;
            }
        }
        ss.erase(a[i-k]);
    }
    printf("%d\n",ans);
    return 0;
}