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>
using namespace std;
const int N = 2000005, M = 8000005;
int a[N], q[N], f[N], v[M], mx[M];
void add(int x, int l, int r, int ll, int rr) {
  if (r < ll || l > rr) return;
  if (ll <= l && r <= rr) {
    v[x]++;
    return;
  }
  if (v[x]) {
    v[2 * x] += v[x];
    v[2 * x + 1] += v[x];
    v[x] = 0;
  }
  int m = (l + r) / 2;
  add(x * 2, l, m, ll, rr);
  add(x * 2 + 1, m + 1, r, ll, rr);
  mx[x] = max(mx[x * 2] + v[x * 2], mx[x * 2 + 1] + v[x * 2 + 1]);
}
int qry(int x, int l, int r, int ll, int rr) {
  if (r < ll || l > rr) return 0;
  if (ll <= l && r <= rr) return v[x] + mx[x];
  if (v[x]) {
    v[2 * x] += v[x];
    v[2 * x + 1] += v[x];
    v[x] = 0;
  }
  int m = (l + r) / 2;
  return max(qry(2 * x, l, m, ll, rr), qry(2 * x + 1, m + 1, r, ll, rr));
}
int main() {
  int n, k, h = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = n; i >= 1; i--) {
    while (h > 0 && a[q[h - 1]] <= a[i]) {
      f[q[h - 1]] = i;
      h--;
    }
    q[h++] = i;
  }
  for (int i = 1; i <= n; i++) f[i] = max(f[i], i - k);
  for (int i = 1; i <= n; i++) {
    add(1, 1, n, f[i] + 1, i);
    if (i >= k) printf("%d\n", qry(1, 1, n, i - k + 1, i));
  }
  return 0;
}
