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:
#include <bits/stdc++.h>
using namespace std;

int solve() {
    int N, K;
    cin >> N >> K;
    vector<int> P(N);
    for ( int i = 0; i < N; i++ ) {
        cin >> P[i];
    }
    int c = 0;
    set<int> s;
    int ans = 1;
    s.insert(P[0]);
    for ( int i = 1; i < K; i++ ) {
        if ( P[i] < P[i-1] ) c++;
        s.insert(P[i]);
    }
    bool had_zero = (c==0);
    for ( int i = K; i < N; i++ ) {
        if ( P[i] < P[i-1] ) c++;
        if ( P[i-K+1] < P[i-K] ) c--;
        s.erase(P[i-K]);
        int mn = *s.begin();
        int mx = *s.rbegin();
        bool no = P[i-K] < mn && mx < P[i];
        s.insert(P[i]);
        if ( !no ) {
            if ( c > 0 ) {
                ans++;
            } else if ( c==0 && !had_zero ) {
                ans++;
                had_zero = true;
            }
        }
    }
    
    return ans;
}

int main() {
    auto ans = solve();
    cout << ans << "\n";
    return 0;
}