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[4000005], d[4000005], ad[4000005], last[4000005], inf = 1e9;
void build(int k, int l, int r) {
  d[k] = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(k * 2, l, mid);
  build(k * 2 + 1, mid + 1, r);
}
int query(int k, int l, int r, int x, int y) {
  if (x > y) return 0;
  if (x <= l && y >= r) return d[k];
  int mid = (l + r) >> 1;
  if (y <= mid)
    return query(k * 2, l, mid, x, y);
  else if (x > mid)
    return query(k * 2 + 1, mid + 1, r, x, y);
  else
    return max(query(k * 2, l, mid, x, mid),
               query(k * 2 + 1, mid + 1, r, mid + 1, y));
}
void update(int k, int l, int r, int x, int y) {
  if (l == r) {
    d[k] = max(d[k], y);
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    update(k * 2, l, mid, x, y);
  else
    update(k * 2 + 1, mid + 1, r, x, y);
  d[k] = max(d[k * 2], d[k * 2 + 1]);
}
void build1(int k, int l, int r) {
  d[k] = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build1(k * 2, l, mid);
  build1(k * 2 + 1, mid + 1, r);
}
void update1(int k, int l, int r, int x, int y, int z) {
  if (x > y) return;
  if (x <= l && y >= r) {
    ad[k] += z;
    d[k] += z;
    return;
  }
  int mid = (l + r) >> 1;
  if (y <= mid)
    update1(k * 2, l, mid, x, y, z);
  else if (x > mid)
    update1(k * 2 + 1, mid + 1, r, x, y, z);
  else
    update1(k * 2, l, mid, x, mid, z),
        update1(k * 2 + 1, mid + 1, r, mid + 1, y, z);
  d[k] = max(d[k * 2], d[k * 2 + 1]) + ad[k];
}
int query1(int k, int l, int r, int x, int y, int add) {
  if (x <= l && y >= r) return add + d[k];
  int mid = (l + r) >> 1;
  if (y <= mid)
    return query1(k * 2, l, mid, x, y, add + ad[k]);
  else if (x > mid)
    return query1(k * 2 + 1, mid + 1, r, x, y, add + ad[k]);
  else
    return max(query1(k * 2, l, mid, x, mid, add + ad[k]),
               query1(k * 2 + 1, mid + 1, r, mid + 1, y, add + ad[k]));
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    last[i] = query(1, 1, n, a[i], n);
    update(1, 1, n, a[i], i);
  }
  build1(1, 1, n);
  for (int i = 1; i < m; i++) update1(1, 1, n, last[i] + 1, i, 1);
  for (int i = m; i <= n; i++) {
    update1(1, 1, n, last[i] + 1, i, 1);
    printf("%d ", query1(1, 1, n, i - m + 1, i, 0));
  }
  return 0;
}
