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 maxn = 2e5 + 10;
int p[maxn];

set<int>st;

int main() 
{
    int n, k; cin >> n >> k;
    for (int i = 0; i < n; i++)
        cin >> p[i];
    int inc = 1;
    int tmp = 0;
    for (int i = 0; i < k; i++) {
        st.insert(p[i]);
        if (i > 0) 
            inc = p[i] > p[i - 1] ? inc + 1 : 1;
    }
    int ans = inc >= k ? 0 : 1;
    tmp = inc >= k ? 1 : tmp;
    for (int i = k; i < n; i++) {
        inc = p[i] > p[i - 1] ? inc + 1 : 1;
        
        tmp = inc >= k ? 1 : tmp;
        int j = i - k;
        st.erase(p[j]);
        if (p[i] > *st.rbegin() && p[j] < *st.begin()) {
        } else ans += inc >= k ? 0 : 1;
        st.insert(p[i]);
    }
    cout << ans + tmp << endl;
}