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 rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, k;
	cin >> n >> k;
	vector<int> p(n);
	rep(i,n) cin >> p[i];
	set<int> st;
	for(int i = 1; i < k; ++i){
		st.insert(p[i]);
	}
	int ans = 1;
	rep(i,n-k){
		if(!(p[i] < *st.begin() && *st.rbegin() < p[i+k])) ++ans;
		st.erase(p[i+1]);
		st.insert(p[i+k]);
	}
	int cnt = 1, dup = 0;
	rep(i,n-1){
		if(p[i] < p[i+1]) ++cnt;
		else cnt = 1;
		if(cnt == k) ++dup;
	}
	if(dup > 0) ans -= dup - 1;
	cout << ans << endl;
	return 0;
}
