Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
using P = pair<int, int>;
const int MX = 200005;
int n, k, p, b, e, b2, e2, ans;
bool s;
P deq[MX], deq2[MX];
int main(){
	scanf("%d%d", &n, &k);
	ans = n-k+1;
	rep(i,n){
		bool ok = true;
		scanf("%d", &p);
		
		while(b2 < e2 && deq2[e2-1].second < p) e2--;
		if(e2 > b2) ok = false;
		deq2[e2++] = P(i,p);
		if(deq2[b2].first + k == i) b2++;
		
		while(b < e && deq[e-1].second > p) e--;
		deq[e++] = P(i,p);
		if(deq[b].first + k == i) b++;
		else ok = false;
		
		if(ok || (s && e-b == k)) ans--;
		if(e-b == k) s = true;
	}
	printf("%d\n", ans);
}