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
/*
 * Author       : YangDavid
 * Created Time : 2019年09月21日 星期六 21时32分40秒
 */

#include<bits/stdc++.h>
#define rep(i, n) for(int i = 1; i <= n; ++i)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 222222;
int n, k, p[maxn], ge[maxn];

int main() {
    scanf("%d%d", &n, &k);
    rep(i, n) scanf("%d", &p[i]);
    for(int i = 2; i <= n; ++i)
        ge[i] = p[i] > p[i - 1], ge[i] += ge[i - 1];
    set<int> st;

    int same = ge[k] == k - 1, tot = 1 - same;
    rep(i, k) st.insert(p[i]);
    for(int i = k + 1; i <= n; ++i) {
        bool ne = false;
        if(*st.begin() != p[i - k]) ne = true;
        st.erase(p[i - k]);
        if(*st.rbegin() > p[i]) ne = true;
        st.insert(p[i]);
        if(ne && ge[i] - ge[i - k + 1] == k - 1) same |= 1, ne = false;
        tot += ne;
    }
    printf("%d\n", tot + same);
    return 0;
}
