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;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 1e6 + 5;
const double eps = 1e-9;
int a[MAXN], st[MAXN], en[MAXN];
vector<int> adj[MAXN];
int t = 0, tree[4 * MAXN], lazy[4 * MAXN];
void prop(int l, int r, int k) {
  if (l != r) {
    lazy[k * 2] += lazy[k];
    lazy[k * 2 + 1] += lazy[k];
  }
  lazy[k] = 0;
}
int qry(int l, int r, int k, int a, int b) {
  tree[k] += lazy[k];
  prop(l, r, k);
  if (r < a || b < l) return 0;
  if (a <= l && r <= b) return tree[k];
  int m = (l + r) / 2;
  return max(qry(l, m, k * 2, a, b), qry(m + 1, r, k * 2 + 1, a, b));
}
void upd(int l, int r, int k, int a, int b, int v) {
  tree[k] += lazy[k];
  prop(l, r, k);
  if (r < a || b < l) return;
  if (a <= l && r <= b) {
    tree[k] += v;
    lazy[k] += v;
    prop(l, r, k);
    return;
  }
  int m = (l + r) / 2;
  upd(l, m, k * 2, a, b, v);
  upd(m + 1, r, k * 2 + 1, a, b, v);
  tree[k] = max(tree[k * 2], tree[k * 2 + 1]);
}
void add(int u, int v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void dfs(int u, int p) {
  st[u] = t++;
  for (int v : adj[u])
    if (v != p) dfs(v, u);
  en[u] = t;
}
int ind[MAXN];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < (n); i++) scanf("%d", &(a[i]));
  deque<int> pq;
  vector<int> root;
  for (int i = n - 1; i >= 0; i--) {
    while (!pq.empty() && a[i] >= a[pq.back()]) pq.pop_back();
    if (!pq.empty())
      add(i, pq.back());
    else
      root.push_back(i);
    pq.push_back(i);
  }
  for (int i = 0; i < n; i++) sort(adj[i].begin(), adj[i].end());
  reverse(root.begin(), root.end());
  for (int u : root) dfs(u, -1);
  for (int i = 0; i < n; i++) {
    upd(0, n - 1, 1, st[i], en[i] - 1, 1);
    if (i >= k - 1) {
      printf("%d ", qry(0, n - 1, 1, 0, n - 1));
      upd(0, n - 1, 1, st[i - k + 1], en[i - k + 1] - 1, -1);
    }
  }
  return 0;
}
