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 + 10;
int n, k, v[maxn], sta[maxn], id[maxn], topt, root, l[maxn], top;
struct da {
  int lc, rc, tag, ma;
} a[4 * maxn];
inline void updata(int n) { a[n].ma = max(a[a[n].lc].ma, a[a[n].rc].ma); }
void build_tree(int &n, int l, int r) {
  n = ++topt;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build_tree(a[n].lc, l, mid);
  build_tree(a[n].rc, mid + 1, r);
}
void push_down(int n) {
  if (!a[n].tag) return;
  a[a[n].lc].tag += a[n].tag;
  a[a[n].lc].ma += a[n].tag;
  a[a[n].rc].tag += a[n].tag;
  a[a[n].rc].ma += a[n].tag;
  a[n].tag = 0;
}
void tree_add(int n, int L, int R, int l, int r, int kk) {
  if (L == l && R == r) {
    a[n].tag += kk;
    a[n].ma += kk;
    return;
  }
  int mid = (L + R) >> 1;
  push_down(n);
  if (r <= mid)
    tree_add(a[n].lc, L, mid, l, r, kk);
  else if (l >= mid + 1)
    tree_add(a[n].rc, mid + 1, R, l, r, kk);
  else
    tree_add(a[n].lc, L, mid, l, mid, kk),
        tree_add(a[n].rc, mid + 1, R, mid + 1, r, kk);
  updata(n);
}
int qury(int n, int L, int R, int l, int r) {
  if (L == l && R == r) return a[n].ma;
  int mid = (L + R) >> 1;
  push_down(n);
  if (r <= mid)
    return qury(a[n].lc, L, mid, l, r);
  else if (l >= mid + 1)
    return qury(a[n].rc, mid + 1, R, l, r);
  else
    return max(qury(a[n].lc, L, mid, l, mid),
               qury(a[n].rc, mid + 1, R, mid + 1, r));
  updata(n);
}
int main() {
  scanf("%d%d", &n, &k);
  build_tree(root, 1, n);
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (int i = 1; i <= n; i++) {
    while (top && sta[top] < v[i]) top--;
    l[i] = id[top] + 1;
    sta[++top] = v[i];
    id[top] = i;
  }
  for (int i = 1; i <= n; i++) {
    tree_add(root, 1, n, l[i], i, 1);
    if (i >= k)
      printf("%d%c", qury(root, 1, n, i - k + 1, i), (i == n ? '\n' : ' '));
  }
  return 0;
}
