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;
using ll = long long int;
using ull = unsigned long long int;
using dd = double;
using ldd = long double;
using si = short int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
vector<vector<int>> gr;
void build_gr(vector<int>& arr) {
  int n = arr.size();
  gr.resize(n);
  vector<int> ind;
  for (int i = 0; i < n; ++i) {
    while (ind.size() && arr[ind.back()] < arr[i]) {
      gr[i].push_back(ind.back());
      ind.pop_back();
    }
    ind.push_back(i);
  }
}
int _size = 1 << 20;
vector<ll> fir, las, tree(_size << 1, 0);
int sz = 0;
vector<bool> was(_size);
vector<int> _upd(_size << 1);
void dfs(int v) {
  was[v] = 1;
  fir[v] = sz;
  ++sz;
  for (int i : gr[v]) {
    dfs(i);
  }
  las[v] = sz - 1;
}
void build_eu() {
  fir.resize(gr.size());
  las.resize(gr.size());
  for (int i = gr.size() - 1; i >= 0; --i)
    if (!was[i]) dfs(i);
}
void push(int v) {
  if (v >= _size) return;
  _upd[v << 1] += _upd[v];
  _upd[v << 1 | 1] += _upd[v];
  tree[v << 1] += _upd[v];
  tree[v << 1 | 1] += _upd[v];
  _upd[v] = 0;
}
void upd(int v, int l, int r, int fl, int fr, ll pl) {
  if (r <= l || r <= fl || fr <= l) return;
  if (fl <= l && r <= fr) {
    if (pl == -1) {
      tree[v] = _upd[v] = -1e9;
      return;
    }
    _upd[v] += pl;
    tree[v] += pl;
    return;
  }
  push(v);
  upd(v << 1, l, (r + l) >> 1, fl, fr, pl);
  upd(v << 1 | 1, (r + l) >> 1, r, fl, fr, pl);
  tree[v] = max(tree[v << 1], tree[v << 1 | 1]);
}
ll get(int v, int l, int r, int fl, int fr) {
  if (r <= l || r <= fl || fr <= l) return -1e9;
  if (fl <= l && r <= fr) return tree[v];
  push(v);
  return max(get(v << 1, l, (r + l) >> 1, fl, fr),
             get(v << 1 | 1, (r + l) >> 1, r, fl, fr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) cin >> arr[i];
  build_gr(arr);
  build_eu();
  for (int i = 0; i < k; ++i) {
    upd(1, 0, _size, fir[i], las[i] + 1, 1);
  }
  for (int i = k; i < n; ++i) {
    cout << tree[1] << " ";
    upd(1, 0, _size, fir[i], las[i] + 1, 1);
    upd(1, 0, _size, fir[i - k], las[i - k] + 1, -1);
  }
  cout << tree[1];
  0;
  0;
  return 0;
}
