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>
const int N = 1000200;
const long long mod = 1000000007;
using namespace std;
int n;
int k;
int a[N];
int r[N];
int d[N];
int p[N];
int dip[N];
multiset<int> s[N];
vector<int> g[N];
multiset<int> res;
int get(int x) { return x == p[x] ? x : p[x] = get(p[x]); }
int best(int x) {
  if (s[x].empty()) {
    return 1;
  }
  return dip[x] - *s[x].begin() + 1;
}
void add(int x) {
  if (r[x]) {
    g[r[x]].push_back(x);
  }
  s[x].insert(dip[x]);
  for (int y : g[x]) {
    if (x - y >= k) {
      continue;
    }
    int py = get(y);
    res.erase(res.find(best(py)));
    if (s[py].size() > s[x].size()) {
      swap(s[x], s[py]);
    }
    for (int z : s[py]) {
      s[x].insert(z);
    }
    s[py].clear();
    p[py] = x;
  }
  res.insert(best(x));
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  vector<int> v;
  for (int i = n; i >= 1; i--) {
    p[i] = i;
    while (!v.empty() && a[i] >= a[v.back()]) {
      v.pop_back();
    }
    if (!v.empty()) {
      r[i] = v.back();
      dip[i] = dip[r[i]] - 1;
    }
    v.push_back(i);
  }
  for (int i = 1; i <= k; i++) {
    add(i);
  }
  printf("%d ", *(--res.end()));
  for (int i = 1; i <= n - k; i++) {
    int x = get(i);
    res.erase(res.find(best(x)));
    if (x != i) {
      s[x].erase(s[x].find(dip[i]));
      res.insert(best(x));
    }
    add(i + k);
    printf("%d ", *(--res.end()));
  }
}
