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 mxN = 1e6;
int n, k, a[mxN], l[mxN];
struct dsu {
  int p[mxN], ld[mxN], lb[mxN];
  void init() {
    iota(p, p + n, 0);
    iota(lb, lb + n, 0);
  }
  int find(int x) { return x == p[x] ? x : (p[x] = find(p[x])); }
  void join(int x, int y) {
    x = find(x), y = find(y);
    p[x] = y;
    lb[y] = lb[x];
    ld[y] += ld[x];
  }
  void upd(int i, int j) {
    i = find(i);
    ++ld[i];
    if (j < n) --ld[j];
    while (ld[i] >= 0 && lb[i]) join(lb[i] - 1, i);
  }
  void upd2(int i) {
    if (i < k) return;
    int j = find(i - k + 1);
    while (lb[j] > i - k) join(lb[j] - 1, j);
  }
  int qry() { return ld[find(0)]; }
} d;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  d.init();
  for (int i = 0; i < n; ++i) {
    d.upd2(i);
    l[i] = i - 1;
    while (l[i] >= 0 && a[l[i]] < a[i]) {
      l[i] = l[l[i]];
    }
    d.upd(l[i] + 1, i + 1);
    if (i >= k - 1) cout << d.qry() << " ";
  }
  return 0;
}
