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;
int a[1000010], pre[1000010];
int mx[1000010 << 2], flag[1000010 << 2];
void push_up(int u) { mx[u] = max(mx[(u << 1)], mx[((u << 1) | 1)]); }
void push_down(int u) {
  if (!flag[u]) return;
  mx[(u << 1)] += flag[u];
  mx[((u << 1) | 1)] += flag[u];
  flag[(u << 1)] += flag[u];
  flag[((u << 1) | 1)] += flag[u];
  flag[u] = 0;
}
void update(int u, int L, int R, int st, int ed) {
  if (st >= R || ed <= L) return;
  if (st <= L && ed >= R) {
    mx[u]++;
    flag[u]++;
    return;
  }
  push_down(u);
  update((u << 1), L, (L + R >> 1), st, ed);
  update(((u << 1) | 1), (L + R >> 1), R, st, ed);
  push_up(u);
}
int calc(int u, int L, int R, int st, int ed) {
  if (st >= R || ed <= L) return 0;
  if (st <= L && ed >= R) return mx[u];
  push_down(u);
  return max(calc((u << 1), L, (L + R >> 1), st, ed),
             calc(((u << 1) | 1), (L + R >> 1), R, st, ed));
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    int st = i - 1;
    while (st && a[st] < a[i]) st = pre[st];
    pre[i] = st;
    update(1, 1, n + 1, pre[i] + 1, i + 1);
    if (i >= k) cout << calc(1, 1, n + 1, i - k + 1, i + 1) << ' ';
  }
}
