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 N=200005;
int n,k,a[N],ans=1,bl[N];
set<int> st;
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<=k;i++) st.insert(a[i]);
    int l=1;
    for(int i=2;i<=n;i++)
    {
        if(a[i]<a[i-1]) l=i;
        else if(i-l+1>=k) bl[i]=1;
    }
    int used=bl[k];
    for(int i=k+1;i<=n;i++)
    {
        int be=*st.begin();
        if(bl[i]&&used)
        {
            st.erase(a[i-k]); st.insert(a[i]);
            continue;
        }
        if(be!=a[i-k]||st.lower_bound(a[i])!=st.end()) 
        {
            ans++;
        }
        if(bl[i]) used=1;
        st.erase(a[i-k]); st.insert(a[i]);
    }
    cout<<ans<<endl;
    return 0;
}