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 = 1e6 + 7;
int a[N], c[N];
struct o_O {
  int max, lazy;
} f[N * 4];
void pushdown(int x) {
  if (f[x].max == -1) return;
  f[x * 2].max += f[x].lazy;
  f[x * 2 + 1].max += f[x].lazy;
  f[x * 2].lazy += f[x].lazy;
  f[x * 2 + 1].lazy += f[x].lazy;
  f[x].lazy = 0;
}
int query(int x, int l, int r, int left, int right) {
  if (l > right || r < left) return 0;
  if (left <= l && right >= r) return f[x].max;
  pushdown(x);
  return max(query(x * 2, l, (l + r) / 2, left, right),
             query(x * 2 + 1, (l + r) / 2 + 1, r, left, right));
}
void update(int x, int l, int r, int left, int right) {
  if (l > right || r < left) return;
  if (left <= l && right >= r) {
    if (f[x].max != -1) {
      f[x].max += 1;
      f[x].lazy += 1;
    }
    return;
  }
  pushdown(x);
  update(x * 2, l, (l + r) / 2, left, right);
  update(x * 2 + 1, (l + r) / 2 + 1, r, left, right);
  f[x].max = max(f[x * 2].max, f[x * 2 + 1].max);
}
void update0(int x, int l, int r, int index) {
  if (l > index || r < index) return;
  if (l == r) {
    f[x].max = -1;
    return;
  }
  pushdown(x);
  update0(x * 2, l, (l + r) / 2, index);
  update0(x * 2 + 1, (l + r) / 2 + 1, r, index);
  f[x].max = max(f[x * 2].max, f[x * 2 + 1].max);
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int l = 0, r = 0;
  c[0] = 0;
  for (int i = 1; i <= n; i++) {
    while (r > l && a[i] > a[c[r]]) r--;
    update(1, 1, n, c[r] + 1, i);
    c[++r] = i;
    if (i >= k) {
      printf("%d ", query(1, 1, n, i - k + 1, i));
      update0(1, 1, n, i - k + 1);
    }
  }
  return 0;
}
