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 ll          long long
#define ld          long double

const ll mod=1e9+7;

void solve()  
{
	ll n, k; cin>>n>>k; 
	ll a[n];
	for(ll i=0; i<n; i++) cin>>a[i];
	if(n==k) { cout<<1<<endl; return; }
	vector<ll>go(n, 0);
	ll ans=0;
	for(ll i=1; i<n; i++)
	{
		if(a[i]>a[i-1]) go[i]=go[i-1]+1;
		else { go[i]=0; }
		if(go[i] >= k-1) ans=1;
	}
	set<ll>v;
	for(ll i=0; i<k; i++) v.insert(a[i]);
	if(go[k-1]!=k-1) ans++;
	for(ll i=k; i<n; i++)
	{
		v.insert(a[i]);
		ll st=*v.begin();
		ll en=*v.rbegin();
		if(st != a[i-k] or en != a[i])
		{
			if(go[i] < k-1) ans++;
		}
		v.erase(a[i-k]);
	}
	cout<<ans<<endl;
}
 
int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);  
	ll T = 1;
	//cin>>T; 
	while(T--)
	    solve();
	return 0;
}
