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;
const int N = 2e5+100;
#define fr(i,low,up) for(register int i = low;i <= up; ++i)
#define nfr(i,low,up) for(register int i = low;i >= up; --i)

int n,k;
int p[N];
int sum[N];
bool flag = false;
set<int> s1;
int main(){
   int ans = 0;
   scanf("%d%d",&n,&k);
   fr(i,1,n){
      scanf("%d",p+i);
      sum[i] = p[i] > p[i-1];
   }
   fr(i,1,n)sum[i] += sum[i-1];
   fr(i,1,k)s1.insert(p[i]);

   if(sum[k]-sum[1] == k-1)flag = true;
   ans = 1;
   fr(i,1,n-k){
      s1.insert(p[i+k]);
      if(p[i] != *s1.begin() || p[i+k] != *--s1.end()){
         if(sum[i+k]-sum[i+1] != k-1)++ans;
         else if(!flag && sum[i+k]-sum[i+1] == k-1)++ans,flag = true;
      }
      s1.erase(p[i]);
   }
   printf("%d\n",ans);
   return 0;
}
