Problem: 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

Solution (in CPP):

Solution:
// ※※※ 解答不能 ※※※
// 正解者のソースを学習する(一部改変).
// hbi1998氏
// https://atcoder.jp/contests/agc038/submissions/7638873
#include <bits/stdc++.h>
using namespace std;
const int N = 220000;

int a[N];

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