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 n, k;
int num[1000001];
int L[1000001];
int sum_tree[1000001 * 3];
int max_tree[1000001 * 3];
void update(int a, int b, int i = 1, int start = 1, int end = n + 1) {
  if (start == a && end == b) {
    max_tree[i] += 1;
    sum_tree[i] += 1;
    return;
  }
  int mid = (start + end) / 2;
  if (a < mid) update(a, min(b, mid), i * 2, start, mid);
  if (b > mid) update(max(a, mid), b, i * 2 + 1, mid, end);
  max_tree[i] = max(max_tree[i * 2], max_tree[i * 2 + 1]) + sum_tree[i];
}
int query(int a, int b, int i = 1, int start = 1, int end = n + 1) {
  if (start == a && end == b) return max_tree[i];
  int mid = (start + end) / 2;
  int ans = 0;
  if (a < mid) ans = max(ans, query(a, min(b, mid), i * 2, start, mid));
  if (b > mid) ans = max(ans, query(max(a, mid), b, i * 2 + 1, mid, end));
  return ans + sum_tree[i];
}
int main() {
  cin >> n >> k;
  num[0] = 1 << 30;
  for (int i = 1; i <= n; ++i) {
    cin >> num[i];
  }
  for (int i = 1; i <= n; ++i) {
    int idx = i - 1;
    while (num[i] > num[idx]) {
      idx = L[idx];
    }
    L[i] = idx;
  }
  for (int i = 1; i <= n; ++i) {
    update(L[i] + 1, i + 1);
    if (i >= k) {
      cout << query(i - k + 1, i + 1) << ' ';
    }
  }
  cout << '\n';
}
