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>
#define ALL(A) (A).begin(),(A).end()
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)

const ll mod = 1e9+7;
const ll INF = -1*((1LL<<63)+1);
const int inf = -1*((1<<31)+1);

using namespace std;

int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N,K;
    cin >> N >> K;
    int ans = 1;
    vector<int> a(N),b(N);
    rep(i,N)cin >> a[i];
    b[0] = 1;
    for(int i=1;i<N;i++){
        if(a[i]>a[i-1])b[i] = b[i-1] + 1;
        else b[i] = 1;
    }
    set<int> st;
    rep(i,K)st.insert(a[i]);
    bool f = K==b[K-1];
    for(int i=K;i<N;i++){
        st.insert(a[i]);//i番目の数を入れる
        int mx = *st.rbegin();
        int mn = *st.begin();
        st.erase(a[i-K]);//一番左の数を捨てる
        
        if(mx==a[i] && mn == a[i-K])continue;
        if(b[i]!=K)ans++;
        else if(!f){
            ans++,f=1;
        } 
    }
    cout << ans << endl;
}
