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 <vector>
#include <algorithm>
#include <string>
#include <set>

using namespace std;
typedef long long ll;
ll mod = 1000000007;

int main() {
    int n,k;
    scanf("%d %d",&n,&k);
    vector<int> p(n,0),t(n-k+1,0),sm(n-k+1,0);
    for (int i = 0; i < n; i++) scanf("%d",&p[i]);
    int cnt = 0, j = 1;
    for (int i = 0; i < n - k + 1; i++) {
        while (j < i + k) {
            if (p[j-1]>p[j]) cnt++;
            j++;
        }
        t[i] = cnt;
        if (p[i] > p[i+1]) cnt--;
    }
    set<int> s;
    int  l = 0;
    for (int i = 0; i < n - k; i++) {
        while (l < i + k + 1) s.insert(p[l++]);
        if (*s.begin() == p[i] && *prev(s.end()) == p[l - 1]) sm[i + 1] = 1;
        s.erase(p[i]);
    }
    int ans = 0, z = 0;
    for (int i = 0; i < n - k + 1; i++) if (!sm[i]) {
        if (t[i]) ans++;
        else z++;
    }
    ans += bool(z);
    printf("%d\n",ans);
}