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 c = 1000002, f = 21, cc = 4194304, kk = cc / 2;
int n, k, t[c], cnt, be[c], ki[c], kezd[cc], veg[cc], maxi[cc], lp[cc];
vector<int> sz[c], p;
bool v[c];
void dfs(int a) {
  v[a] = true;
  cnt++, be[a] = cnt;
  for (int x : sz[a]) {
    dfs(x);
  }
  cnt++, ki[a] = cnt;
}
void add(int a, int l, int r, int e) {
  if (kezd[a] > r || veg[a] < l) return;
  if (l <= kezd[a] && veg[a] <= r) {
    lp[a] += e;
    return;
  }
  int x = 2 * a, y = 2 * a + 1;
  add(x, l, r, e), add(y, l, r, e);
  maxi[a] = max(maxi[x] + lp[x], maxi[y] + lp[y]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    while (p.size() > 0 && t[i] > t[p.back()]) {
      sz[i].push_back(p.back());
      p.pop_back();
    }
    p.push_back(i);
  }
  for (int i = n; i; i--) {
    if (!v[i]) {
      dfs(i);
    }
  }
  for (int i = kk; i < cc; i++) kezd[i] = i - kk, veg[i] = i - kk;
  for (int i = kk - 1; i; i--) kezd[i] = kezd[2 * i], veg[i] = veg[2 * i + 1];
  for (int i = 1; i < k; i++) {
    add(1, be[i], ki[i], 1);
  }
  for (int i = k; i <= n; i++) {
    add(1, be[i], ki[i], 1);
    cout << maxi[1] + lp[1] << " ";
    int s = i - k + 1;
    add(1, be[s], ki[s], -1);
  }
  cout << "\n";
  return 0;
}
