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 (auto &p : P) cin >> p;
    vector<bool> E(N);
    int eql = 0, tot = 0;
    int ans = N - K + 1;
    for (int i = 0; i < N - 1; i++) {
        if (P[i] < P[i + 1]) eql++;
        tot++;
        if (tot == K - 1 && eql == K - 1) {
            E[i + 2 - K] = 1;
        }
        if (i + 2 - K >= 0) {
            if (P[i + 2 - K] < P[i + 3 - K]) eql--;
            tot--;
        }
    }
    set<int> S;
    for (int i = 0; i < N; i++) {
        if (S.size() < K) {
            S.insert(P[i]);
        } else {
            if (P[i - K] == *begin(S) && P[i] > *prev(end(S))) {
                if (!E[i - K]) ans--;
            }
            S.insert(P[i]);
            S.erase(P[i - K]);
        }
    }
    int cnt = 0;
    for (auto e : E) if (e) cnt++;
    cout << ans - max(0, cnt - 1) << endl;
    return 0;
}
