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 N = 1e6 + 5;
int n, k, a[N], l[N];
long long seg[N << 2], lazy[N << 2];
stack<int> s;
void shift(int id) {
  if (lazy[id]) {
    int lc = id * 2, rc = lc | 1;
    seg[lc] += lazy[id];
    seg[rc] += lazy[id];
    lazy[lc] += lazy[id];
    lazy[rc] += lazy[id];
    lazy[id] = 0;
  }
}
void add(int l, int r, int val, int id = 1, int b = 0, int e = n) {
  if (l >= e || b >= r) return;
  if (l <= b && e <= r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  shift(id);
  int mid = (b + e) >> 1, lc = id * 2, rc = lc | 1;
  add(l, r, val, lc, b, mid);
  add(l, r, val, rc, mid, e);
  seg[id] = max(seg[lc], seg[rc]);
}
long long get(int l, int r, int id = 1, int b = 0, int e = n) {
  if (l >= e || b >= r) return 0;
  if (l <= b && e <= r) return seg[id];
  shift(id);
  int mid = (b + e) >> 1, lc = id * 2, rc = lc | 1;
  return max(get(l, r, lc, b, mid), get(l, r, rc, mid, e));
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    l[i] = -1;
  }
  for (int i = 0; i < n; i++) {
    while (!s.empty() && a[s.top()] < a[i]) s.pop();
    if (s.size()) l[i] = s.top();
    s.push(i);
  }
  for (int i = 0; i < k - 1; i++) add(l[i] + 1, i + 1, 1);
  for (int i = k - 1; i < n; i++) {
    add(l[i] + 1, i + 1, 1);
    cout << get(max(0, i - k + 1), i + 1) << " ";
  }
  cout << "\n";
}
