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 <iostream>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;

#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
#define FOR(i, m, n) for(int i = (int)(m); i < (int)(n); ++i)

int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> P(N);
    REP(i, N) cin >> P[i];

    // P[i:i+K] が元々昇順に並んでいるような i の個数を求める
    int length = 1, num = 0;
    REP(i, N-1){
        if(P[i] < P[i+1]){ 
            ++length;
            if(length == K) ++num;
        }else length = 1;
    }

    int ans = 1;
    // 連結成分の個数を求める
    set<int> s;
    REP(i, K) s.insert(P[i]);
    FOR(i, K, N){
        s.erase(P[i-K]);
        if(P[i-K] > *s.begin() or P[i] < *s.rbegin()) ++ans;
        s.insert(P[i]);
    }
    cout << ans-max(0, num-1) << endl;

    return 0;
}