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=2e5+10;
int a[maxn];
int main(){
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;++i)cin>>a[i];
    int m=0;
    int ans=n-k+1;
    for(int i=2;i<=n;){
        if(a[i]>a[i-1]){
            int len=0;
            while(i<=n&&a[i]>a[i-1]){
                len++;
                i++;
            }
            if(len>=k-1)m++;
        }
        else
            i++;
    }
    set<int>s;
    int l=1,r=k;
    for(int i=l;i<=r;++i)s.insert(a[i]);
    while(l<=r&&r<=n){
        r++;
        if(r>n)break;
        s.insert(a[r]);
        if(*(s.rbegin())==a[r]&&*(s.begin())==a[l])ans--;
        s.erase(a[l]);
        l++;
    }
    if(!m)cout<<ans<<endl;
    else cout<<ans-m+1<<endl;
}
