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 mi = -20000000;
const int maxn = 1000050;
struct node {
  int l;
  int r;
  int lc;
  int rc;
  int mx;
  int tag;
} s[2 * maxn];
struct edge {
  int t;
  edge *next;
} e[maxn];
int a[maxn], p[maxn];
int deep[maxn], num[maxn];
int ans[maxn];
int n, k;
edge *head[maxn];
int etot, dfn, tot;
void ade(int f, int t) {
  etot++;
  e[etot].t = t;
  e[etot].next = head[f];
  head[f] = &e[etot];
}
stack<int> st;
int dfs(int x, int dep) {
  p[x] = ++dfn;
  deep[x] = dep;
  num[x] = 1;
  for (edge *i = head[x]; i; i = i->next) {
    num[x] += dfs(i->t, dep + 1);
  }
  return num[x];
}
int build(int l, int r) {
  int now = tot++;
  s[now].l = l;
  s[now].r = r;
  if (l == r) return now;
  int mid = (l + r) >> 1;
  s[now].rc = build(mid + 1, r);
  s[now].lc = build(l, mid);
  return now;
}
void pushdown(int id) {
  node now = s[id];
  s[now.rc].tag += now.tag;
  s[now.lc].tag += now.tag;
  s[now.rc].mx += now.tag;
  s[now.lc].mx += now.tag;
  s[id].tag = 0;
}
void change(int id, int l, int r, int x) {
  if (l <= s[id].l && s[id].r <= r) {
    s[id].mx += x;
    s[id].tag += x;
    return;
  }
  int mid = (s[id].l + s[id].r) >> 1;
  pushdown(id);
  if (l <= mid) change(s[id].lc, l, r, x);
  if (r > mid) change(s[id].rc, l, r, x);
  s[id].mx = max(s[s[id].lc].mx, s[s[id].rc].mx);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = n; i >= 1; i--) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (st.empty())
      ade(n + 1, i);
    else
      ade(st.top(), i);
    st.push(i);
  }
  dfs(n + 1, 0);
  build(1, n + 1);
  change(0, 1, n + 1, mi);
  for (int i = n; i >= n - k + 2; i--) change(0, p[i], p[i], deep[i] - mi);
  for (int i = n - k + 1; i >= 1; i--) {
    change(0, p[i], p[i], deep[i] - mi);
    ans[i] = s[0].mx;
    change(0, p[i + k - 1], p[i + k - 1] + num[i + k - 1] - 1, -1);
  }
  for (int i = 1; i <= n - k + 1; i++) printf("%d ", ans[i]);
  return 0;
}
