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 <cstdio>
#include <set>

using namespace std;

int a[200000];
int p[200000];

int main() {
    int n, k, ans = 1, c = 0, i;
    set <int> s;
    
    scanf("%d %d", &n, &k);
    
    for (i = 0; i < n; i++) scanf("%d", &p[i]);
    
    a[0] = 1;
    for (i = 1; i < n; i++) {
        if (p[i] > p[i - 1]) {
            a[i] = a[i - 1] + 1;
        } else {
            a[i] = 1;
        }
    }
    
    if (a[k - 1] == k) c = 1;
    
    for (i = 0; i < k; i++) s.insert(p[i]);
    
    for (i = k; i < n; i++) {
        s.insert(p[i]);
        
        if (a[i] >= k) {
            if (c == 0) ans++;
            c = 1;
        } else if (*s.begin() != p[i - k] || *s.rbegin() != p[i]) {
            ans++;
        }
        
        s.erase(p[i - k]);
    }
    
    printf("%d\n", ans);
    
    return 0;
}
