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 int long long
#define rep(i,n) for(int i=0;i<n;i++)

int N, K;
int A[200010];
bool mem[200010];
set<int> S;

signed main()
{
    cin >> N >> K;
    rep(i, N)cin >> A[i];

    int l = 0;
    rep(i, N)
    {
        if(i && A[i - 1] > A[i])l = i;
        if(i - l + 1 >= K)mem[i - K + 1] = true;
    }

    int ans = 0;
    rep(i, N)if(mem[i])ans = 1;

    rep(i, K)S.insert(A[i]);

    if(!mem[0])ans++;
    rep(i, N - K)
    {
        S.insert(A[K + i]);
        if(!(*(S.begin()) == A[i] && *(--S.end()) == A[K + i])
        && !mem[i + 1])ans++;
        S.erase(A[i]);
    }

    cout << ans << endl;

    return 0;
}
