Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main() {
    int n, k; cin >> n >> k;
    vector<int> p(n);
    for(auto& x : p) cin >> x;

    vector<int> dp(n);
    for(int i = 1; i < n; ++i) {
        if(p[i - 1] < p[i]) {
            dp[i] = dp[i - 1] + 1;
        }
    }

    set<int> s;
    int ans = 0, sorted = 0;
    if(dp[k - 1] == k - 1) {
        sorted = 1;
    } else {
        ans = 1;
    }
    for(int i = 0; i < k; ++i) {
        s.insert(p[i]);
    }
    for(int i = k; i < n; ++i) {
        bool add = *s.begin() != p[i - k];
        s.insert(p[i]);
        s.erase(p[i - k]);
        add |= *prev(s.end()) != p[i];
        if(add) {
            if(dp[i] >= k - 1) {
                sorted = 1;
            } else {
                ans += 1;
            }
        }
    }
    ans += sorted;

    cout << ans << endl;
}