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;
struct edge {
  int to, next;
} e[1000010];
struct Seg {
  int c, add;
} tr[4000010];
int n, K, a[1000010], head[1000010], siz[1000010], dfn[1000010], cnt, tot, top,
    S[1000010];
inline void add(int u, int v) {
  cnt++;
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
inline void dfs(int u) {
  dfn[u] = ++tot;
  siz[u] = 1;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    dfs(v);
    siz[u] += siz[v];
  }
}
inline void push_up(int p) { tr[p].c = max(tr[p << 1].c, tr[p << 1 | 1].c); }
inline void push_down(int p) {
  if (tr[p].add == 0) return;
  tr[p << 1].c += tr[p].add;
  tr[p << 1 | 1].c += tr[p].add;
  tr[p << 1].add += tr[p].add;
  tr[p << 1 | 1].add += tr[p].add;
  tr[p].add = 0;
}
inline void update(int p, int l, int r, int L, int R, int val) {
  if (l >= L && r <= R) {
    tr[p].c += val;
    tr[p].add += val;
    return;
  }
  push_down(p);
  int mid = (l + r) >> 1;
  if (L <= mid) update(p << 1, l, mid, L, R, val);
  if (R > mid) update(p << 1 | 1, mid + 1, r, L, R, val);
  push_up(p);
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  n++;
  a[n] = n;
  for (int i = 1; i <= n; i++) {
    while (top && a[S[top]] < a[i]) {
      add(i, S[top]);
      top--;
    }
    S[++top] = i;
  }
  dfs(n);
  for (int i = 1; i <= K; i++) update(1, 1, n, dfn[i], dfn[i] + siz[i] - 1, 1);
  printf("%d ", tr[1].c);
  for (int i = K + 1; i < n; i++) {
    update(1, 1, n, dfn[i - K], dfn[i - K] + siz[i - K] - 1, -1);
    update(1, 1, n, dfn[i], dfn[i] + siz[i] - 1, 1);
    printf("%d ", tr[1].c);
  }
  return 0;
}
