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 = 1e6 + 10;
int arr[N];
int par[N];
vector<int> adj[N];
int start[N], ending[N];
int ctr = 0;
int tree[4 * N] = {0};
int lazy[4 * N] = {0};
int ans[N];
inline int maxi(int a, int b) { return a > b ? a : b; }
void dfs(int node) {
  start[node] = ctr++;
  for (int x : adj[node]) {
    dfs(x);
  }
  ending[node] = ctr - 1;
}
inline void upd(int node, int l, int r) {
  if (lazy[node] == 0 or l == r) {
    lazy[node] = 0;
    return;
  }
  tree[2 * node] += lazy[node];
  lazy[2 * node] += lazy[node];
  tree[2 * node + 1] += lazy[node];
  lazy[2 * node + 1] += lazy[node];
  lazy[node] = 0;
}
void update(int node, int l, int r, int left, int right, int val) {
  upd(node, l, r);
  if (right < l or r < left) {
    return;
  }
  if (l == r) {
    tree[node] += val;
    return;
  }
  if (left <= l and r <= right) {
    tree[node] += val;
    lazy[node] += val;
    return;
  }
  int m = (l + r) / 2;
  update(node * 2, l, m, left, right, val);
  update(2 * node + 1, m + 1, r, left, right, val);
  tree[node] = maxi(tree[2 * node], tree[2 * node + 1]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  par[n] = 0;
  stack<int> s;
  s.push(n);
  for (int i = n - 1; i >= 1; i--) {
    while (!s.empty() and arr[s.top()] <= arr[i]) s.pop();
    if (s.empty()) {
      par[i] = 0;
    } else
      par[i] = s.top();
    s.push(i);
  }
  for (int i = 1; i <= n; i++) {
    adj[par[i]].push_back(i);
  }
  dfs(0);
  for (int i = 1; i <= k; i++) {
    update(1, 0, n, start[i], ending[i], 1);
  }
  ans[k] = tree[1];
  for (int i = k + 1; i <= n; i++) {
    update(1, 0, n, start[i], ending[i], 1);
    update(1, 0, n, start[i - k], ending[i - k], -1);
    ans[i] = tree[1];
  }
  for (int i = k; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
