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 MAXN = 1e6 + 5;
struct seg {
  int tr[MAXN << 2], lazy[MAXN << 2];
  inline void push_up(int rt) { tr[rt] = max(tr[rt << 1], tr[rt << 1 | 1]); }
  inline void push_down(int rt) {
    if (lazy[rt]) {
      tr[rt << 1] += lazy[rt];
      tr[rt << 1 | 1] += lazy[rt];
      lazy[rt << 1] += lazy[rt];
      lazy[rt << 1 | 1] += lazy[rt];
      lazy[rt] = 0;
    }
  }
  void build(int l, int r, int rt) {
    tr[rt] = lazy[rt] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
  }
  void update(int L, int R, int val, int l, int r, int rt) {
    if (L <= l && r <= R) {
      tr[rt] += val;
      lazy[rt] += val;
      return;
    }
    int mid = (l + r) >> 1;
    push_down(rt);
    if (L <= mid) update(L, R, val, l, mid, rt << 1);
    if (mid < R) update(L, R, val, mid + 1, r, rt << 1 | 1);
    push_up(rt);
  }
} se;
stack<int> st;
int a[MAXN], in[MAXN], out[MAXN], tim = 0;
vector<int> E[MAXN];
void dfs(int now) {
  in[now] = ++tim;
  for (int v : E[now]) {
    dfs(v);
  }
  out[now] = tim;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  n++;
  a[n] = 1e9;
  for (int i = 1; i <= n; i++) {
    while (!st.empty() && a[st.top()] < a[i]) {
      E[i].push_back(st.top());
      st.pop();
    }
    st.push(i);
  }
  dfs(n);
  se.build(1, n, 1);
  for (int i = 1; i < n; i++) {
    se.update(in[i], out[i], 1, 1, n, 1);
    if (i > k) se.update(in[i - k], out[i - k], -1, 1, n, 1);
    if (i >= k) printf("%d ", se.tr[1]);
  }
  puts("");
  return 0;
}
