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;

void solve() {
    int n, k; cin >> n >> k;
    vector<int> nums(n);
    vector<int> ord(n, 1);
    set<int> st;
    bool flag = 0;
    for (int i = 0; i < n; ++i) cin >> nums[i];
    for (int i = n-2; i >= 0; --i) {
        if (nums[i] < nums[i+1]) ord[i] = ord[i+1] + 1;
        else ord[i] = 1;
        if (ord[i] >= k) flag = true;
    }
    for (int i=0; i < k; ++i) {
        st.insert(nums[i]);
    }
    int ret = ord[0] >= k ? 0 : 1;// ord[0] < k;
    for (int i=k; i < n; ++i) {
        st.insert(nums[i]);
        int mx = *st.rbegin(), mi = *st.begin();
        if (nums[i] != mx || nums[i-k] != mi){
            if (ord[i-k+1] < k) {
                ret++;
            }
        }
        st.erase(nums[i-k]);
    }
    cout << (ret + flag) << endl;
}

int main() {
    int n; n = 1;
    while (n--) {
        solve();
    }
}