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 n, m, a[1200000], stk[1200000], dfn[1200000], head[1200000], tot, clk,
    fst[1200000], lst[1200000], seg[1200000 << 2], fa[1200000], mx[1200000],
    dep[1200000];
struct edge {
  int v, nxt;
} e[1200000];
multiset<int> se;
int gfa(int x) { return x == fa[x] ? x : fa[x] = gfa(fa[x]); }
void add(int x, int y) {
  e[++tot].v = y;
  e[tot].nxt = head[x];
  head[x] = tot;
}
void dfs(int u) {
  dfn[u] = ++clk;
  fst[u] = lst[u] = clk;
  for (int i = head[u], v; i; i = e[i].nxt) {
    dep[v = e[i].v] = dep[u] + 1;
    dfs(v);
    lst[u] = clk;
  }
}
void mdf(int x, int l, int r, int t, int k) {
  if (l == r) {
    seg[x] = k;
    return;
  }
  int mid = (l + r) >> 1;
  if (t <= mid)
    mdf((x << 1), l, mid, t, k);
  else
    mdf((x << 1 | 1), mid + 1, r, t, k);
  seg[x] = max(seg[(x << 1)], seg[(x << 1 | 1)]);
}
int qry(int x, int l, int r, int L, int R) {
  if (l >= L && r <= R) return seg[x];
  int mid = (l + r) >> 1, tmp = 0;
  if (L <= mid) tmp = max(tmp, qry((x << 1), l, mid, L, R));
  if (R > mid) tmp = max(tmp, qry((x << 1 | 1), mid + 1, r, L, R));
  return tmp;
}
void del(int x) { se.erase(se.find(mx[x])); }
void renew(int x) {
  se.insert(mx[x] = qry(1, 1, n + 1, fst[x], lst[x]) - dep[x] + 1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  int tp = 0;
  for (int i = 1; i <= n; ++i) {
    for (; tp && a[i] > a[stk[tp]]; --tp) add(i, stk[tp]);
    stk[++tp] = i;
  }
  for (; tp; --tp) add(0, stk[tp]);
  dfs(0);
  for (int i = 1; i <= n; ++i) fa[i] = i;
  for (int i = 1, j = 1; i <= n - m + 1; ++i) {
    if (i - 1) {
      mdf(1, 1, n + 1, dfn[i - 1], 0);
      del(gfa(i - 1));
      renew(gfa(i - 1));
    }
    for (; j < i + m; ++j) {
      mdf(1, 1, n + 1, dfn[j], dep[j]);
      for (int k = head[j]; k; k = e[k].nxt) {
        del(e[k].v);
        fa[e[k].v] = j;
      }
      renew(j);
    }
    printf("%d ", *(--se.end()));
  }
  puts("");
  return 0;
}
