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 rep(i,n) for(int i=0; i<(n); i++)
using P = pair<int, int>;
const int MX = 200005;
int n, k, p[MX], l, r, ans, s = -1;
P qu[MX];
bool a[MX];
void push(int t, bool k){
	while(l < r && (k ^ (qu[r-1].first > p[t]))) r--;
	qu[r++] = P(p[t],t);
}
int main(){
	scanf("%d%d", &n, &k);
	rep(i,n){
		scanf("%d", p+i);
		push(i,0);
		if(qu[l].second < i-k+1) l++;
		if(qu[l].second == i-k+1) a[i-k+1] = true;
		if(r-l == k){
			if(s != -1 && s+k <= i) ans--;
			s = i;
		}
	}
	l=r=0;
	for(int i = n-1; i > 0; i--){
		push(i,1);
		if(i+k-1 < qu[l].second) l++;
		if(i+k-1 == qu[l].second && a[i-1]) ans--;
	}
	printf("%d\n", ans+n-k+1);
}