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;
template <typename G>
struct triple {
  G first, second, T;
};
struct segment_tree {
  vector<int> st;
  vector<int> lazy;
  segment_tree(int n) : st(2 * n), lazy(2 * n) {}
  inline int id(int b, int e) { return (b + e - 1) | (b != e - 1); }
  void prop(int l, int r) {
    int cur = id(l, r);
    st[cur] += lazy[cur];
    if (l != r - 1) {
      int mid = (l + r + 1) >> 1;
      lazy[id(l, mid)] += lazy[cur];
      lazy[id(mid, r)] += lazy[cur];
    }
    lazy[cur] = 0;
  }
  void sum(int l, int r, int li, int ri, int v) {
    if (li == ri) return;
    int cur = id(l, r);
    if (lazy[cur]) prop(l, r);
    if (l >= ri || r <= li) return;
    if (li <= l && r <= ri) {
      lazy[cur] += v;
      prop(l, r);
      return;
    }
    int mid = (l + r + 1) >> 1;
    sum(l, mid, li, ri, v);
    sum(mid, r, li, ri, v);
    st[cur] = max(st[id(l, mid)], st[id(mid, r)]);
  }
  int query(int l, int r, int li, int ri) {
    int cur = id(l, r);
    if (lazy[cur]) prop(l, r);
    if (l >= ri || r <= li) return 0;
    if (li <= l && r <= ri) return st[cur];
    int mid = (l + r + 1) >> 1;
    return max(query(0, mid, li, ri), query(mid, r, li, ri));
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> v(n);
  for (auto &i : v) cin >> i;
  vector<int> p(n + 1, -1);
  stack<int> s;
  for (int i = n - 1; i >= 0; i--) {
    while (!s.empty() && v[s.top()] <= v[i]) s.pop();
    if (!s.empty())
      p[i] = s.top();
    else
      p[i] = n;
    s.push(i);
  }
  ++n;
  vector<vector<int>> tree(n);
  for (int i = 0; i < n - 1; i++) tree[p[i]].push_back(i);
  vector<int> l(n), r(n);
  int t = 0;
  function<void(int)> dfs = [&](int c) {
    l[c] = t++;
    for (auto a : tree[c]) dfs(a);
    r[c] = t;
  };
  dfs(n - 1);
  segment_tree st(n);
  for (int i = 0; i < k; i++) st.sum(0, n, l[i], r[i], 1);
  for (int i = k; i < n - 1; i++) {
    cout << st.query(0, n, 0, n) << " ";
    st.sum(0, n, l[i - k], r[i - k], -1);
    st.sum(0, n, l[i], r[i], 1);
  }
  cout << st.query(0, n, 0, n) << '\n';
  return 0;
}
