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;
using ll = long long;

int main() {
    int N, K;
    cin >> N >> K;
    vector<int>P(N);
    for (int i = 0; i < N; ++i) {
        cin >> P[i];
    }

    // 一番左が最小, 追加されるのが最大ならソートされない
    set<int> se;
    for (int i = 0; i < K; ++i) {
        se.insert(P[i]);
    }

    ll ans = N - K + 1;
    int cnt = 1;
    int def = 0;
    set<int>flag;
    for (int i = 1; i < N; ++i) {
        if(P[i-1] < P[i]) {
            cnt++;
        } else{
            cnt = 1;
        }
        if(cnt >= K){
            flag.insert(i);
            def++;
            cnt--;
        }
    }
    if(def) ans = ans + 1 - def;

    for (int i = K; i < N; ++i) {
        if(*se.begin() == P[i-K] && se.upper_bound(P[i]) == se.end()){
            if(! flag.count(i)) ans--;
        }
        se.erase(P[i-K]);
        se.insert(P[i]);
    }
    cout << ans;
    return 0;
}