For some array c, let's denote a greedy subsequence as a sequence of indices p_1, p_2, ..., p_l such that 1 ≤ p_1 < p_2 < ... < p_l ≤ |c|, and for each i ∈ [1, l - 1], p_{i + 1} is the minimum number such that p_{i + 1} > p_i and c[p_{i + 1}] > c[p_i].

You are given an array a_1, a_2, ..., a_n. For each its subsegment of length k, calculate the length of its longest greedy subsequence.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10^6) — the length of array a and the length of subsegments.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — array a.

Output

Print n - k + 1 integers — the maximum lengths of greedy subsequences of each subsegment having length k. The first number should correspond to subsegment a[1..k], the second — to subsegment a[2..k + 1], and so on.

Examples

Input


6 4
1 5 2 5 3 6


Output


2 2 3 


Input


7 6
4 5 2 5 3 6 6


Output


3 3 

Note

In the first example: 

  * [1, 5, 2, 5] — the longest greedy subsequences are 1, 2 ([c_1, c_2] = [1, 5]) or 3, 4 ([c_3, c_4] = [2, 5]). 
  * [5, 2, 5, 3] — the sequence is 2, 3 ([c_2, c_3] = [2, 5]). 
  * [2, 5, 3, 6] — the sequence is 1, 2, 4 ([c_1, c_2, c_4] = [2, 5, 6]). 



In the second example: 

  * [4, 5, 2, 5, 3, 6] — the longest greedy subsequences are 1, 2, 6 ([c_1, c_2, c_6] = [4, 5, 6]) or 3, 4, 6 ([c_3, c_4, c_6] = [2, 5, 6]). 
  * [5, 2, 5, 3, 6, 6] — the subsequence is 2, 3, 5 ([c_2, c_3, c_5] = [2, 5, 6]). 
#include <bits/stdc++.h>
const int inf = 1e9;
using namespace std;
int v[1000005], st[1000005], t[1000005], aux[1000005];
int findt(int nod) {
  if (nod == t[nod]) return nod;
  return t[nod] = findt(t[nod]);
}
int main() {
  int n, k, i;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (i = 1; i <= n; i++) {
    st[i] = i - 1;
    while (st[i] && v[st[i]] < v[i]) st[i] = st[st[i]];
  }
  int j = 1, sum = 0;
  for (i = 1; i <= n; i++) {
    t[i] = i;
    int poz = findt(st[i] + 1);
    if (poz <= j) sum++;
    aux[poz]++;
    aux[i + 1]--;
    if (poz) t[poz] = poz - 1;
    if (i >= k) printf("%d ", sum), sum += aux[++j];
  }
  return 0;
}
