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;
vector<int> v[2000006];
bitset<2000006> is;
int in[2000006], out[2000006], t;
struct SEG {
  struct Node {
    int tag, mx;
  } node[4 * 2000006];
  inline void pull(int id) {
    node[id].mx = max(node[id << 1].mx, node[id << 1 | 1].mx);
  }
  inline void push(int id) {
    int tmp = node[id].tag;
    if (tmp == 0) return;
    node[id].tag = 0;
    node[id << 1].tag += tmp;
    node[id << 1].mx += tmp;
    node[id << 1 | 1].tag += tmp;
    node[id << 1 | 1].mx += tmp;
  }
  inline void update(int l, int r, int ll, int rr, int k, int id) {
    if (l > rr || r < ll) return;
    push(id);
    if (l >= ll && r <= rr) {
      node[id].tag += k;
      node[id].mx += k;
      return;
    }
    update(l, ((l + r) >> 1), ll, rr, k, id << 1);
    update(((l + r) >> 1) + 1, r, ll, rr, k, id << 1 | 1);
    pull(id);
  }
} seg;
int n, k, a[2000006], la[2000006];
inline void DFS(int now) {
  is[now] = 1;
  in[now] = ++t;
  for (int i : v[now]) {
    DFS(i);
  }
  out[now] = t;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  stack<int> st;
  for (int i = n; i >= 1; --i) {
    while (st.size() && a[st.top()] <= a[i]) st.pop();
    if (st.size()) {
      v[st.top()].push_back(i);
      la[i]++;
    }
    st.push(i);
  }
  for (int i = 1; i <= n; ++i) {
    if (la[i] == 0) DFS(i);
  }
  for (int i = 1; i <= n; i++) {
    if (i - k >= 1) {
      seg.update(1, t, in[i - k], out[i - k], -1, 1);
    }
    seg.update(1, t, in[i], out[i], 1, 1);
    if (i >= k) cout << seg.node[1].mx << ' ';
  }
  return 0;
}
