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;
const int maxn = 2e5+10;

int n, k;
int niz[maxn];
bool bio[maxn];
set<int> s;

int main() {
    memset(bio, false, sizeof bio);
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++)
        scanf("%d", niz+i);

    if (k == 1) {printf("1\n"); return 0;}

    int sol = n - k + 1;
    for (int i = 1; i < k; i++)
        s.insert(niz[i]);

    for (int i = k; i < n; i++) {
        if (niz[i] > *s.rbegin() && niz[i - k] < *s.begin()) sol--, bio[i] = true;
        s.erase(niz[i - k + 1]);
        s.insert(niz[i]);
    }

    int tren = 0;
    int kol = 0;
    for (int i = 1; i < n; i++) {
        if (niz[i] > niz[i - 1]) tren++;
        else tren = 0;

        if (tren >= k - 1 && !bio[i]) kol++;
    }
    sol -= max(0, kol - 1);
    printf("%d", sol);
    return 0;
}
