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], p[1000005];
basic_string<int> e[1000005], st;
int l[1000005], r[1000005], dub[1000005], lrt;
void dfs(int x) {
  l[x] = ++lrt;
  for (int y : e[x]) {
    dub[y] = dub[x] + 1;
    dfs(y);
  }
  r[x] = lrt;
}
const int MAXN = 1 << 20;
int b[2 * MAXN];
void update(int x, int v) {
  x += MAXN;
  b[x] = v;
  while (x > 1) {
    x >>= 1;
    b[x] = max(b[2 * x], b[2 * x + 1]);
  }
}
int query(int l, int r, int x = 1, int xl = 0, int xr = MAXN - 1) {
  if (r < xl || xr < l) return 0;
  if (l <= xl && xr <= r) return b[x];
  int xm = (xl + xr) >> 1;
  return max(query(l, r, 2 * x, xl, xm), query(l, r, 2 * x + 1, xm + 1, xr));
}
bool alive[1000005];
multiset<int> root_vals;
set<pair<int, int>> root_ls;
void add_node(int x) {
  for (int y : e[x]) {
    if (alive[y]) {
      int v = query(l[y], r[y]) - dub[y];
      root_vals.erase(root_vals.find(v));
      root_ls.erase({l[y], y});
    }
  }
  update(l[x], dub[x]);
  int v = query(l[x], r[x]) - dub[x];
  root_vals.insert(v);
  root_ls.insert({l[x], x});
  alive[x] = 1;
}
void remove_node(int x) {
  auto it = prev(root_ls.upper_bound({l[x], 123123123}));
  int root = it->second;
  int v = query(l[root], r[root]) - dub[root];
  root_vals.erase(root_vals.find(v));
  update(l[x], 0);
  if (x != root) {
    v = query(l[root], r[root]) - dub[root];
    root_vals.insert(v);
  } else {
    root_ls.erase({l[x], x});
  }
  alive[x] = 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  n++;
  a[n] = n;
  for (int i = 1; i <= n; i++) {
    while (st.size() && a[st.back()] < a[i]) {
      p[st.back()] = i;
      st.pop_back();
    }
    st += i;
  }
  for (int i = 1; i < n; i++) e[p[i]] += i;
  dfs(n);
  for (int i = 1; i <= k; i++) add_node(i);
  cout << *root_vals.rbegin() + 1 << ' ';
  for (int i = 2; i <= n - k; i++) {
    add_node(i + k - 1);
    remove_node(i - 1);
    cout << *root_vals.rbegin() + 1 << ' ';
  }
  cout << '\n';
}
