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 int N_MAX = 1e6 + 1;
int n, k, a[N_MAX], E[2 * N_MAX], fE[N_MAX], lE[N_MAX];
stack<int> st;
vector<int> G[N_MAX];
bool visited[N_MAX];
int Elen = 0;
void DFS(int u) {
  E[Elen++] = u;
  visited[u] = true;
  for (int v : G[u]) {
    if (visited[v]) continue;
    DFS(v);
    E[Elen++] = u;
  }
}
int arr[2 * N_MAX], F[2 * N_MAX], lazy[2 * N_MAX];
void pass_lazy(int a, int b) {
  int k = (a + b) / 2;
  F[k] += lazy[k];
  if (a == k)
    arr[a] += lazy[k];
  else
    lazy[(a + k) / 2] += lazy[k];
  if (k + 1 == b)
    arr[b] += lazy[k];
  else
    lazy[(b + k + 1) / 2] += lazy[k];
  lazy[k] = 0;
}
int add(int a, int b, int i, int j, int val) {
  if (a == b) {
    arr[a] += val;
    return arr[a];
  }
  int k = (a + b) / 2;
  if (a == i && b == j) {
    lazy[k] += val;
    return F[k] + lazy[k];
  }
  if (lazy[k] != 0) pass_lazy(a, b);
  if (j <= k)
    F[k] =
        max(add(a, k, i, j, val),
            (k + 1 == b ? arr[b] : F[(b + k + 1) / 2] + lazy[(b + k + 1) / 2]));
  else if (k + 1 <= i)
    F[k] = max((a == k ? arr[a] : F[(a + k) / 2] + lazy[(a + k) / 2]),
               add(k + 1, b, i, j, val));
  else
    F[k] = max(add(a, k, i, k, val), add(k + 1, b, k + 1, j, val));
  return F[k];
}
int query(int a, int b, int i, int j) {
  if (a == b) return arr[a];
  int k = (a + b) / 2;
  if (lazy[k] != 0) pass_lazy(a, b);
  if (a == i && b == j) return F[k];
  if (j <= k) return query(a, k, i, j);
  if (k + 1 <= i) return query(k + 1, b, i, j);
  return max(query(a, k, i, k), query(k + 1, b, k + 1, j));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  a[n] = 1e6 + 1;
  st.push(n);
  for (int i = n - 1; 0 <= i; i--) {
    while (a[st.top()] <= a[i]) st.pop();
    G[i].push_back(st.top());
    G[st.top()].push_back(i);
    st.push(i);
  }
  DFS(n);
  for (int i = 0; i < Elen; i++) lE[E[i]] = i;
  for (int i = Elen - 1; 0 <= i; i--) fE[E[i]] = i;
  for (int i = 0; i < n; i++) {
    add(0, Elen - 1, fE[i], lE[i], 1);
    if (k <= i) add(0, Elen - 1, fE[i - k], lE[i - k], -1);
    if (k - 1 <= i) cout << query(0, Elen - 1, 0, Elen - 1) << " ";
  }
  cout << endl;
  return 0;
}
