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 MAXN = 1e6 + 10;
int n, k;
int a[MAXN];
void read_input() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
}
int mi[MAXN];
vector<int> graph[MAXN];
int in[MAXN], out[MAXN];
int timer;
void dfs(int v) {
  in[v] = timer++;
  for (auto x : graph[v]) {
    dfs(x);
  }
  out[v] = timer++;
}
const int TSIZE = (1 << 23);
int tree[TSIZE], lazy[TSIZE];
void push(int k, int l, int r) {
  if (lazy[k]) {
    tree[k] += lazy[k];
    if (l != r) {
      lazy[k * 2] += lazy[k];
      lazy[k * 2 + 1] += lazy[k];
    }
    lazy[k] = 0;
  }
}
void update(int k, int l, int r, int i, int j, int val) {
  push(k, l, r);
  if (r < i || l > j || l > r) return;
  if (r <= j && l >= i) {
    tree[k] += val;
    if (l != r) {
      lazy[k * 2] += val;
      lazy[k * 2 + 1] += val;
    }
    return;
  }
  int middle = (l + r) >> 1;
  update(k * 2, l, middle, i, j, val);
  update(k * 2 + 1, middle + 1, r, i, j, val);
  tree[k] = max(tree[k * 2], tree[k * 2 + 1]);
}
void solve() {
  stack<pair<int, int>> st;
  for (int i = n; i >= 1; i--) {
    pair<int, int> temp = {i, a[i]};
    while (!st.empty() && st.top().second <= a[i]) {
      st.pop();
    }
    int ans = n + 1;
    if (!st.empty()) ans = st.top().first;
    st.push(temp);
    graph[ans].push_back(i);
  }
  dfs(n + 1);
  for (int i = 1; i <= k - 1; i++) {
    update(1, 0, timer - 1, in[i], out[i], 1);
  }
  for (int i = k; i <= n; i++) {
    update(1, 0, timer - 1, in[i], out[i], 1);
    push(1, 0, timer - 1);
    cout << tree[1] << " ";
    update(1, 0, timer - 1, in[i - k + 1], out[i - k + 1], -1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  read_input();
  solve();
  return 0;
}
