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;
int head[1000010], n, k, a[1000010], tree[4000010], add[4000010],
    st[1000010][2], t = 0, o = 0, size[1000010], iid[1000010], x = 0;
struct edge {
  int to, link;
} e[1000010];
void add_edge(int u, int v) { e[++o].to = v, e[o].link = head[u], head[u] = o; }
void dfs(int u) {
  iid[u] = ++x;
  size[u] = 1;
  for (int i = head[u]; i; i = e[i].link) {
    dfs(e[i].to);
    size[u] += size[e[i].to];
  }
}
void pushdown(int t) {
  tree[t << 1] += add[t];
  add[t << 1] += add[t];
  tree[t << 1 | 1] += add[t];
  add[t << 1 | 1] += add[t];
  add[t] = 0;
}
void pushup(int t) { tree[t] = max(tree[t << 1], tree[t << 1 | 1]); }
void add_tree(int ll, int rr, int l, int r, int t, int c) {
  if (ll <= l && r <= rr) {
    tree[t] += c, add[t] += c;
    return;
  }
  pushdown(t);
  int mid = (l + r) >> 1;
  if (mid >= ll) add_tree(ll, rr, l, mid, t << 1, c);
  if (mid < rr) add_tree(ll, rr, mid + 1, r, t << 1 | 1, c);
  pushup(t);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[++n] = INT_MAX;
  st[++t][0] = a[n], st[t][1] = n;
  for (int i = n - 1; i >= 1; i--) {
    while (st[t][0] <= a[i] && t) t--;
    add_edge(st[t][1], i);
    st[++t][0] = a[i], st[t][1] = i;
  }
  dfs(n);
  for (int i = 1; i <= k; i++)
    add_tree(iid[i], iid[i] + size[i] - 1, 1, n, 1, 1);
  printf("%d ", tree[1]);
  for (int i = k + 1; i < n; i++) {
    add_tree(iid[i], iid[i] + size[i] - 1, 1, n, 1, 1);
    add_tree(iid[i - k], iid[i - k] + size[i - k] - 1, 1, n, 1, -1);
    printf("%d ", tree[1]);
  }
  printf("\n");
}
