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[1000005];
stack<int> s;
int par[1000005];
vector<int> adj[1000006];
int st[1000005], en[1000005];
int in = 1;
int tree[4 * 1000005];
int lazy[4 * 1000005];
void dfs(int i) {
  st[i] = in++;
  for (auto j : adj[i]) dfs(j);
  en[i] = in;
}
void lazify(int s, int e, int i) {
  if (lazy[i]) {
    tree[i] += lazy[i];
    if (s != e) {
      lazy[2 * i] += lazy[i];
      lazy[2 * i + 1] += lazy[i];
    }
    lazy[i] = 0;
  }
}
void upd(int l, int r, int s, int e, int i, int x) {
  lazify(s, e, i);
  if (s > r || e < l || l > r) return;
  if (s >= l && e <= r) {
    lazy[i] += x;
    lazify(s, e, i);
    return;
  }
  int mid = (s + e) >> 1;
  upd(l, r, s, mid, 2 * i, x);
  upd(l, r, mid + 1, e, 2 * i + 1, x);
  tree[i] = max(tree[2 * i], tree[2 * i + 1]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  s.push(n);
  par[n] = n + 1;
  adj[n + 1].push_back(n);
  for (int i = n - 1; i >= 1; i--) {
    while (s.size() && a[s.top()] <= a[i]) s.pop();
    if (s.size())
      par[i] = s.top(), adj[s.top()].push_back(i);
    else
      par[i] = n + 1, adj[n + 1].push_back(i);
    s.push(i);
  }
  dfs(n + 1);
  for (int i = 1; i <= k; i++) upd(st[i], en[i] - 1, 1, in - 1, 1, 1);
  lazify(1, in - 1, 1);
  cout << tree[1] << " ";
  for (int i = k + 1; i <= n; i++) {
    upd(st[i], en[i] - 1, 1, in - 1, 1, 1);
    upd(st[i - k], st[i - k], 1, in - 1, 1, -1000000000);
    lazify(1, in - 1, 1);
    cout << tree[1] << " ";
  }
  cout << '\n';
  return 0;
}
