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 = (int)(1e6) + 322;
const int inf = 0;
const int mod = 1000000007;
const double eps = 1e-9;
int n, k;
int st[N], a[N], id, timer, L[N], R[N];
vector<int> v[N];
void dfs(int x, int lvl = 0, int p = -1) {
  L[x] = timer++;
  for (auto to : v[x]) {
    if (to == p) continue;
    dfs(to, lvl + 1, x);
  }
  R[x] = timer++;
}
struct Node {
  int mx = 0;
  int push = 0;
} t[N * 8];
inline void Merge(Node &acc, Node &left, Node &right) {
  acc.mx = max(left.mx, right.mx);
}
inline void push(int x, int l, int r) {
  if (t[x].push != -inf) {
    t[x].mx += t[x].push;
    if (l != r) {
      t[x + x].push += t[x].push;
      t[x + x + 1].push += t[x].push;
    }
    t[x].push = -inf;
  }
}
void update(int val, int tl, int tr, int x = 1, int l = L[0], int r = R[0]) {
  push(x, l, r);
  if (tl > r || l > tr) return;
  if (tl <= l && r <= tr) {
    t[x].push += val;
    push(x, l, r);
    return;
  }
  int mid = (l + r) >> 1;
  update(val, tl, tr, x + x, l, mid);
  update(val, tl, tr, x + x + 1, mid + 1, r);
  Merge(t[x], t[x + x], t[x + x + 1]);
}
int getmax() {
  push(1, L[0], R[0]);
  return t[1].mx;
}
signed main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    while (id > 0 && a[st[id]] <= a[i]) id--;
    v[i].push_back(st[id]);
    v[st[id]].push_back(i);
    st[++id] = i;
  }
  dfs(0);
  for (int i = 1; i <= k; ++i) {
    update(1, L[i], R[i]);
  }
  vector<int> ans;
  ans.push_back(getmax());
  for (int i = 1; i + k <= n; ++i) {
    update(-1, L[i], R[i]);
    update(1, L[i + k], R[i + k]);
    ans.push_back(getmax());
  }
  reverse(ans.begin(), ans.end());
  for (auto it : ans) cout << it << ' ';
  return 0;
}
