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 maxn = 1e6 + 100;
int a[maxn], n, p[8 * maxn], tree[8 * maxn], k, q[maxn], pre[maxn];
void down(int cha, int con) {
  tree[con] += p[cha];
  p[con] += p[cha];
}
void update(int node, int l, int r, int u, int v) {
  down(node, 2 * node);
  down(node, 2 * node + 1);
  p[node] = 0;
  if (u > r || v < l) return;
  if (u <= l && v >= r) {
    tree[node]++;
    p[node] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  update(node * 2, l, mid, u, v);
  update(node * 2 + 1, mid + 1, r, u, v);
  tree[node] = max(tree[2 * node], tree[2 * node + 1]);
}
int get(int node, int l, int r, int u, int v) {
  down(node, 2 * node);
  down(node, 2 * node + 1);
  p[node] = 0;
  if (u > r || v < l) return 0;
  if (u <= l && v >= r) return tree[node];
  int mid = (l + r) >> 1;
  int res1 = get(2 * node, l, mid, u, v);
  int res2 = get(2 * node + 1, mid + 1, r, u, v);
  return max(res1, res2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int top = 0;
  for (int i = n; i >= 1; i--) {
    while (top && a[q[top]] <= a[i]) pre[q[top]] = i, top--;
    q[++top] = i;
  }
  for (int i = 1; i <= n; i++) {
    update(1, 1, n, pre[i] + 1, i);
    if (i >= k) cout << get(1, 1, n, i - k + 1, i) << " ";
  }
  return 0;
}
