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;
#define REP(i,a) for(int i = 0; i < (a); i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const long long LINF = 1e18;
const long long MOD = 1e9 + 7;

signed main(){
    int n,k;
    cin >> n >> k;
    int p[n];
    REP(i,n){
        cin >> p[i];
    }
    set<int> s;
    REP(i,k){
        s.insert(p[i]);
    }
    int ans = n - k + 1;
    for(int i = k; i < n; i++){
        auto itr = s.end();
        itr--;
        if(p[i - k] == *s.begin() && *itr < p[i]) ans--;
        s.erase(p[i - k]);
        s.insert(p[i]);
    }
    vector<int> sum(n);
    for(int i = 0; i < n - 1; i++){
        sum[i + 1] += sum[i] + (p[i] > p[i + 1]);
    }
    int cnt = 0;
    for(int i = 0; i <= n - k; i++){
        if((i == 0 || p[i - 1] > p[i]) && sum[i + k - 1] - sum[i] == 0) cnt++;
    }
    cnt--;
    cout << ans - max(0, cnt) << endl;
    return 0;
}