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;
#define sz(x) ((int) (x).size())

typedef long long ll;

int N, K;
set<int> S;
vector<int> P;
vector<int> V;
bool srt[200002];

int main(){
	cin >> N >> K;
	P.resize(N);
	int last = -1;
	int sortCnt = 0;
	for(int i=0; i<N; i++){
		cin >> P[i];
		if(P[i] > last){
			sortCnt++;
		}
		else
			sortCnt = 1;
		last = P[i];
		if(sortCnt >= K){
			V.push_back(i);
			srt[i] = true;
		}
	}
	int cnt = 1;
	for(int i=0; i<K; i++){
		S.insert(P[i]);
	}
	for(int i=K; i<N; i++){
		S.insert(P[i]);
		auto it = S.begin();
		if(*it != P[i-K]){
			cnt++;
		}
		else{
			it = S.end();
			it--;
			if(*it != P[i]){
				cnt++;
			}
			else if(srt[i] && srt[i-1])
				cnt++;
		}
		S.erase(P[i-K]);
	}
	ll s = sz(V);
	if(s > 0)
		cnt -= s-1;
	//cout << s << endl;
	cout << cnt << endl;

  return 0;
}
