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, k, i, a[2000005], tin[2000005], tout[2000005], timer;
int t[2000005 * 4], flag[2000005 * 4];
vector<int> g[2000005];
vector<pair<int, int> > st;
void push(int v, int l, int r) {
  if (!flag[v]) return;
  if (l != r) {
    t[v * 2] += flag[v];
    t[v * 2 + 1] += flag[v];
    flag[v * 2] += flag[v];
    flag[v * 2 + 1] += flag[v];
  }
  flag[v] = 0;
}
void upd(int v, int tl, int tr, int l, int r, int x) {
  int m = (tl + tr) / 2;
  push(v, tl, tr);
  if (l > r) return;
  if (tl == l && tr == r) {
    t[v] = max(-1000000000, t[v] + x);
    flag[v] += x;
    return;
  }
  upd(v * 2, tl, m, l, min(r, m), x);
  upd(v * 2 + 1, m + 1, tr, max(l, m + 1), r, x);
  t[v] = max(t[v * 2], t[v * 2 + 1]);
}
void dfs(int v) {
  int i, to;
  tin[v] = timer++;
  for (i = 0; i < g[v].size(); i++) {
    to = g[v][i];
    dfs(to);
  }
  tout[v] = timer++;
}
int main() {
  ios_base ::sync_with_stdio();
  cin.tie(0);
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    while (st.size() > 0 && st.back().first < a[i]) {
      g[i].push_back(st.back().second);
      st.pop_back();
    }
    st.push_back(make_pair(a[i], i));
  }
  while (st.size() != 0) {
    dfs(st.back().second);
    st.pop_back();
  }
  for (i = 0; i < n; i++) {
    if (i - k >= 0) upd(1, 0, 2 * n, tin[i - k], tout[i - k], -n);
    upd(1, 0, 2 * n, tin[i], tout[i], 1);
    if (i >= k - 1) printf("%d ", t[1]);
  }
  return 0;
}
