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];
stack<int> cur;
int nxt[1000005];
vector<int> adj[1000005];
int in_t[1000005], out_t[1000005];
int timer = 0;
void dfs(int x, int p) {
  in_t[x] = timer++;
  for (auto i : adj[x]) {
    if (i == p) continue;
    dfs(i, x);
  }
  out_t[x] = timer - 1;
}
int segtree[4 * 1000005];
int lazy[4 * 1000005];
void push(int x, int l, int r) {
  if (lazy[x] == 0) return;
  segtree[x] += lazy[x];
  if (l != r) {
    lazy[2 * x + 1] += lazy[x];
    lazy[2 * x + 2] += lazy[x];
  }
  lazy[x] = 0;
}
void update(int x, int l, int r, int a, int b, int v) {
  push(x, l, r);
  if (r < a || b < l) return;
  if (a <= l && r <= b) {
    lazy[x] = v;
    push(x, l, r);
    return;
  }
  int m = (l + r) / 2;
  update(2 * x + 1, l, m, a, b, v);
  update(2 * x + 2, m + 1, r, a, b, v);
  segtree[x] = max(segtree[2 * x + 1], segtree[2 * x + 2]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    while (cur.size() > 0 && a[cur.top()] <= a[i]) {
      cur.pop();
    }
    nxt[i] = (cur.size() > 0 ? cur.top() : n);
    cur.push(i);
  }
  for (int i = 0; i < n; i++) {
    adj[i].push_back(nxt[i]);
    adj[nxt[i]].push_back(i);
  }
  dfs(n, -1);
  for (int i = 0; i < k; i++) {
    update(0, 0, n, in_t[i], out_t[i], 1);
  }
  cout << segtree[0] << " ";
  for (int i = 0; i < n - k; i++) {
    update(0, 0, n, in_t[i], out_t[i], -1);
    update(0, 0, n, in_t[i + k], out_t[i + k], 1);
    cout << segtree[0] << " ";
  }
  cout << endl;
}
