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;
#define mp make_pair
const int maxn=2e5;
int a[maxn+5];
int b[maxn+5];
int n,k;
bool f[maxn+5];
set<int> s;
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    for(int i=2;i<=n;++i) if(a[i]>a[i-1]) b[i]=1;
    for(int i=1;i<=n;++i) b[i]+=b[i-1];
    for(int i=k;i<=n;++i) f[i]=true;
    for(int i=k;i<=n;++i)
        if(b[i]-b[i-k+1]==k-1) f[i]=false;
    for(int i=k;i<=n;++i)
        if(b[i]-b[i-k+1]==k-1)
    {
        f[i]=true;
        break;
    }
    f[k]=true;
    for(int i=1;i<=n;++i)
    {
        if(i>k)
        {


            s.erase(a[i-k]);
        //if(i==4) cout<<*s.rbegin()<<" "<<*s.begin()<<endl;
        if(a[i]>*s.rbegin()&&a[i-k]<*s.begin()) f[i]=false;
        }
        s.insert(a[i]);
    }
    int ans=0;
    for(int i=k;i<=n;++i) if(f[i]) ++ans;
    //for(int i=k;i<=n;++i) cout<<i<<" : "<<f[i]<<endl;
    cout<<ans<<endl;
    return 0;
}
