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;
using ll = long long;
using pi = pair<int, int>;

int n, k;
int a[200003];
int b[200003];
set<int> s;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> k;
    for(int i=0;i<n;i++)
        cin >> a[i];
    for(int i=1;i<n;i++)
        b[i] = b[i-1] + (a[i-1] < a[i]);
    int ans = n-k+1;
    int same = 0;
    for(int i=k-1;i<n;i++)
        if(b[i] - b[i-(k-1)] == k-1) same++;
    for(int i=0;i<k;i++)
        s.insert(a[i]);
    for(int i=k;i<n;i++)
    {
        s.erase(a[i-k]);
        if(b[i] - b[i-(k-1)] == k-1)
        {
            s.insert(a[i]);
            continue;
        }
        if(*(s.begin()) > a[i-k] && *(s.rbegin()) < a[i])
        {
            ans--;
        }
        s.insert(a[i]);
    }
    cout << ans - max(same-1, 0);
}