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, a[1000006], nxt[1000006], tin[1000006], tout[1000006];
vector<int> s, g[1000006], ord;
int t[4 * 1000006], d[4 * 1000006];
void dfs(int v) {
  ord.push_back(v);
  tin[v] = ord.size() - 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    dfs(to);
  }
  tout[v] = ord.size() - 1;
}
void push(int v, int l, int r, bool f = 1) {
  if (d[v] == 0) return;
  if (l != r) {
    int m = (l + r) / 2;
    d[v * 2] += d[v];
    d[v * 2 + 1] += d[v];
    if (f) {
      push(v * 2, l, m, 0);
      push(v * 2 + 1, m + 1, r, 0);
    }
  }
  t[v] += d[v];
  d[v] = 0;
}
void add(int v, int l, int r, int i, int j, int val) {
  if (i > j) return;
  push(v, l, r);
  if (l == i && r == j) {
    d[v] += val;
    push(v, l, r);
    return;
  }
  int m = (l + r) / 2;
  add(v * 2, l, m, i, min(m, j), val);
  add(v * 2 + 1, m + 1, r, max(m + 1, i), j, val);
  push(v * 2, l, m);
  push(v * 2 + 1, m + 1, r);
  t[v] = max(t[v * 2], t[v * 2 + 1]);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  a[n + 1] = 1000000007ll;
  s.push_back(n + 1);
  for (int i = n; i >= 1; i--) {
    while (a[s.back()] <= a[i]) s.pop_back();
    nxt[i] = s.back();
    s.push_back(i);
    g[nxt[i]].push_back(i);
  }
  dfs(n + 1);
  for (int i = 1; i <= k; i++) add(1, 0, ord.size() - 1, tin[i], tout[i], 1);
  printf("%d ", t[1]);
  for (int i = k + 1; i <= n; i++) {
    add(1, 0, ord.size() - 1, tin[i], tout[i], 1);
    add(1, 0, ord.size() - 1, tin[i - k], tout[i - k], -1);
    printf("%d ", t[1]);
  }
  cout << endl;
  return 0;
}
