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 ll;
typedef pair<int, int> ii;

const int N = 2e5+100;
int p[N], ok[N];

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL);
	int n, k;cin >> n >> k;
	for(int i=0 ; i<n ; i++)
		cin >> p[i];
	ok[n-1] = 1;
	for(int i=n-2 ; i>=0 ; i--)
		ok[i] = (p[i] < p[i+1] ? ok[i+1] : 0) + 1;
	bool nc = (ok[0] >= k);
	set<int> st;
	int ans=(ok[0] < k);
	for(int i=0 ; i<k-1 ; i++)
		st.insert(p[i]);
	for(int i=1 ; i<=n-k ; i++){
		st.erase(p[i-1]);
		st.insert(p[i+k-2]);
		if(ok[i] >= k){
			nc = true;
			continue;
		}
		if(p[i-1] > *st.begin() || p[i+k-1] < *st.rbegin())
			ans ++;
	}
	cout << ans+nc << "\n";
	return 0;
}














