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 const1 = 998244353;
vector<vector<int> > g;
vector<int> tin, tout;
int timer = 0;
void dfs(int v) {
  tin[v] = timer++;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    dfs(to);
  }
  tout[v] = timer++;
}
vector<int> tree, upd;
void push(int v, int l, int r) {
  if (l == r) {
    return;
  }
  tree[v * 2] += upd[v];
  upd[v * 2] += upd[v];
  tree[v * 2 + 1] += upd[v];
  upd[v * 2 + 1] += upd[v];
  upd[v] = 0;
}
void update(int v, int l, int r, int al, int ar, int val) {
  if (l >= al && r <= ar) {
    upd[v] += val;
    tree[v] += val;
  } else if (l <= ar && r >= al) {
    push(v, l, r);
    update(v * 2, l, (r + l) / 2, al, ar, val);
    update(v * 2 + 1, (r + l) / 2 + 1, r, al, ar, val);
    tree[v] = max(tree[v * 2 + 1], tree[v * 2]);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  tin.resize(n + 1);
  tout.resize(n + 1);
  g.resize(n + 1);
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  stack<int> q;
  for (int i = n - 1; i >= 0; i--) {
    while (q.size() > 0 && a[q.top()] <= a[i]) {
      q.pop();
    }
    if (q.size() == 0) {
      g[0].push_back(i + 1);
    } else {
      g[q.top() + 1].push_back(i + 1);
    }
    q.push(i);
  }
  dfs(0);
  tree.resize(4 * timer);
  upd.resize(4 * timer);
  for (int i = 0; i < n; i++) {
    update(1, 0, timer, tin[i + 1], tout[i + 1], 1);
    if (i >= k) {
      update(1, 0, timer, tin[i + 1 - k], tout[i + 1 - k], -1);
    }
    if (i >= k - 1) {
      cout << tree[1] << " ";
    }
  }
  return 0;
}
