Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

5 3
0 2 1 4 3

Sample Output 1

2

Two permutations can be produced as P after the operation: (0,1,2,4,3) and (0,2,1,3,4).

Sample Input 2

4 4
0 1 2 3

Sample Output 2

1

Sample Input 3

10 4
2 0 1 3 7 5 4 6 8 9

Sample Output 3

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 7;
int a[MAXN];
struct node{int pos, val;};
deque<node>qmin, qmax;
int main()
{
    int n, k; scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= k; i++) qmin.push_back({i, a[i]}), qmax.push_back({i, a[i]});
    int ans = 1;
    for (int i = k + 1; i <= n; i++)
    {
        while (qmin.size() && qmin.front().pos < i - k) qmin.pop_front();
        while (qmin.size() && qmin.back().val >= a[i]) qmin.pop_back();
        qmin.push_back({i, a[i]});

        while (qmax.size() && qmax.front().pos < i - k) qmax.pop_front();
        while (qmax.size() && qmax.back().val <= a[i]) qmax.pop_back();
        qmax.push_back({i, a[i]});

        if (qmin.front().val == a[i - k] && qmax.front().val == a[i]) continue;
        ans++;
    }
    printf("%d\n", ans);
    return 0;
}