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 n,k;

void solve(){
    int i,j;
    string str;
    cin >> n >> k;
    vector<int> p(n);
    for(i = 0;i < n;i++) cin >> p[i];
    set<int> s;
    for(i = 0;i < k;i++) s.insert(p[i]);
    long long ans = 1;
    for(;i < n;i++){
        ans++;
        if(*s.begin() == p[i-k] && *s.rbegin() < p[i]){
            ans--;
        }
        s.erase(p[i-k]);
        s.insert(p[i]);
    }
    vector<int> b(n,0);
    for(i = 0;i < n-1;i++){
        if(p[i] > p[i+1]) b[i] = 1;
    }
    for(i = 0,j = 0;i < n-1;i++){
        if(b[i] == 1){
            j = i+k-1;
        }
        else if(i < j){
            b[i] = 1;
        }
    }
    j = 0;
    int f = 0;
    for(i = k-2;i < n-1;i++){
        if(b[i] == 0 && f == 0){
            j++;
            f = 1;
        }
        if(b[i] == 1) f = 0;
    }
    if(j > 1) ans -= j-1;
    cout << ans << endl;
}

int main(){
    solve();
}
