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;
typedef long long int ll;
typedef pair<ll,ll> mp;
ll inf =  1e9;


int main(){
    ll n,m;
    cin>>n>>m;
    set<ll> st;
    vector<ll> p(n);
    for(ll i=0;i<n;i++){
	cin>>p[i];
    }
    ll res = 1;

    vector<ll> dp(n,1);
    for(ll i=1;i<n;i++) if( p[i-1] < p[i] ){
	dp[i] += dp[i-1];
    }
    bool flag = false;
    for(ll i=0;i<m;i++)st.insert(p[i]);
    for(ll i=1;i<=n-m;i++){
	auto s = st.begin();
	auto t = st.end();
	t--;
	ll mini = *s;
	ll maxi = *t;
	if( (p[i-1]==mini && p[i+m-1] > maxi) || dp[i+m-1] >= m  ){
	    //cout<<p[i-1]<<' '<<p[i+m-1]<<endl;
	}else{
	    res ++;
	}
	if( dp[i + m-1] >= m ) flag =  true;
	st.erase( p[i-1] );
	st.insert( p[i-1+m] );
    }
    if( flag && dp[m-1] <m  ) res ++;
    cout<<res<<endl;

    return 0;
}
