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, tot, num[1000010], L[1000010];
struct node {
  int l, r, mx, sum;
} t[1000010 << 1];
void build(int d, int l, int r) {
  if (l == r) return;
  int mid = l + r >> 1;
  t[d].l = ++tot, build(tot, l, mid), t[d].r = ++tot, build(tot, mid + 1, r);
}
void add(int d, int l, int r, int L, int R, int x) {
  if (L <= l && r <= R) {
    t[d].mx += x, t[d].sum += x;
    return;
  }
  int mid = l + r >> 1;
  if (L <= mid) add(t[d].l, l, mid, L, R, x);
  if (R > mid) add(t[d].r, mid + 1, r, L, R, x);
  t[d].mx = max(t[t[d].l].mx, t[t[d].r].mx) + t[d].sum;
}
int query(int d, int l, int r, int L, int R) {
  if (L <= l && r <= R) return t[d].mx;
  int mid = l + r >> 1, res = 0;
  if (L <= mid) res = max(res, query(t[d].l, l, mid, L, R));
  if (mid < R) res = max(res, query(t[d].r, mid + 1, r, L, R));
  return res + t[d].sum;
}
int main() {
  scanf("%d%d", &n, &m), num[0] = 1000010;
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  for (int i = 1; i <= n; i++) {
    L[i] = i - 1;
    for (; num[i] > num[L[i]]; L[i] = L[L[i]])
      ;
  }
  build(tot = 1, 1, n);
  for (int i = 1; i < m; i++) add(1, 1, n, L[i] + 1, i, 1);
  for (int i = 1, j = m; j <= n; i++, j++)
    add(1, 1, n, L[j] + 1, j, 1), printf("%d ", query(1, 1, n, i, j));
}
