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 a[1000005], st[4 * 1000005], lazy[4 * 1000005] = {};
void update(int start, int end, int l, int r, int dif, int node) {
  if (l > r) return;
  if (lazy[node] != 0) {
    st[node] += lazy[node];
    if (start != end) {
      lazy[2 * node + 1] += lazy[node];
      lazy[2 * node + 2] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start > r or end < l)
    return;
  else if (start >= l and end <= r) {
    st[node] += dif;
    if (start != end) {
      lazy[2 * node + 1] += dif;
      lazy[2 * node + 2] += dif;
    }
    return;
  }
  int mid = (start + end) / 2;
  update(start, mid, l, r, dif, 2 * node + 1);
  update(mid + 1, end, l, r, dif, 2 * node + 2);
  st[node] = max(st[2 * node + 1], st[2 * node + 2]);
}
int queryMin(int start, int end, int l, int r, int node) {
  if (lazy[node] != 0) {
    st[node] += lazy[node];
    if (start != end) {
      lazy[2 * node + 1] += lazy[node];
      lazy[2 * node + 2] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start > r or end < l)
    return 1;
  else if (start >= l and end <= r)
    return st[node];
  else {
    int mid = (start + end) / 2, left = 2 * node + 1, right = left + 1;
    return max(queryMin(start, mid, l, r, left),
               queryMin(mid + 1, end, l, r, right));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  fill(st, st + 4 * 1000005, 1);
  vector<int> root;
  for (int i = 0; i < n; i++) {
    while (root.size() && a[root.back()] < a[i]) {
      root.pop_back();
    }
    int l;
    if (root.size()) {
      l = root.back() + 1;
    } else {
      l = 0;
    }
    root.push_back(i);
    update(0, n - 1, l, i - 1, 1, 0);
    if (i >= k - 1) {
      cout << queryMin(0, n - 1, i - k + 1, i, 0) << " ";
    }
  }
  cout << endl;
  return 0;
}
