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;

int main() {
    int n, k;
    cin >> n >> k;
    vector<int> p(n);
    for (int i = 0; i < n; i++) {
        cin >> p.at(i);
    }

    vector<int> sum(n + 1);
    for (int i = 1; i < n; i++) {
        sum.at(i) = sum.at(i - 1) + (p.at(i - 1) < p.at(i));
    }

    set<int> s;
    for (int i = 0; i < k; i++) {
        s.insert(p.at(i));
    }
    bool is_sorted = false;

    int ans = 0;
    if (sum.at(k - 1) - sum.at(0) == k - 1) is_sorted = true;
    else ans = 1;

    for (int i = k; i < n; i++) {
        s.insert(p.at(i));

        if (sum.at(i) - sum.at(i - k + 1) == k - 1) is_sorted = true;
        else if (!(p.at(i) == *s.rbegin() && p.at(i - k) == *s.begin())) ans++;

        s.erase(p.at(i - k));
    }
    cout << ans + is_sorted << endl;
}