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 s[1000100];
int e[1000100];
int a[1000100];
int cnt = 0, n;
vector<int> pre[1000100];
struct IT {
  int d[4000100];
  int lz[4000100];
  void upd(int num, int l, int r, int x, int y, int val) {
    if (l > r || l > y || r < x) return;
    if (l >= x && r <= y) {
      lz[num] += val;
      d[num] += val;
      return;
    }
    upd(num * 2, l, (l + r) / 2, x, y, val);
    upd(num * 2 + 1, (l + r) / 2 + 1, r, x, y, val);
    d[num] = max(d[num * 2], d[num * 2 + 1]) + lz[num];
  }
  int get(int num, int l, int r, int x, int y) {
    if (l > r || l > y || r < x) return 0;
    if (l >= x && r <= y) {
      return d[num];
    }
    return max(get(num * 2, l, (l + r) / 2, x, y),
               get(num * 2 + 1, (l + r) / 2 + 1, r, x, y)) +
           lz[num];
  }
} it;
void init_dfs(int x) {
  s[x] = ++cnt;
  for (int i = 0; i < pre[x].size(); i++) {
    init_dfs(pre[x][i]);
  }
  e[x] = cnt;
}
void add(int x) { it.upd(1, 1, n, s[x], e[x], 1); }
void rmv(int x) { it.upd(1, 1, n, s[x], e[x], -1); }
stack<int> st;
int main() {
  int k;
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = n; i >= 1; i--) {
    while (st.size() && a[i] >= a[st.top()]) st.pop();
    if (st.size()) {
      pre[st.top()].push_back(i);
    } else {
      pre[n + 1].push_back(i);
    }
    st.push(i);
  }
  n++;
  init_dfs(n);
  for (int i = 1; i <= k; i++) {
    add(i);
  }
  cout << it.get(1, 1, n, 1, n) << " ";
  for (int i = k + 1; i < n; i++) {
    add(i);
    rmv(i - k);
    cout << it.get(1, 1, n, 1, n) << " ";
  }
}
