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 = 1000000 + 10;
int tree[4 * maxn], lazy[4 * maxn], seen[maxn];
int N, K, A[maxn], in[maxn], out[maxn];
vector<int> graph[maxn];
int cnt = 0;
void dfs(int u, int p) {
  seen[u] = 1;
  for (auto &v : graph[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
}
void dfs2(int u, int p) {
  in[u] = ++cnt;
  for (auto &v : graph[u]) {
    if (v == p) continue;
    dfs2(v, u);
  }
  out[u] = cnt;
}
void update(int node, int start, int end, int l, int r, int val) {
  if (lazy[node] != 0) {
    tree[node] += lazy[node];
    if (start != end) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start > end || start > r || end < l) return;
  if (start >= l && end <= r) {
    tree[node] += val;
    if (start != end) {
      lazy[node * 2] += val;
      lazy[node * 2 + 1] += val;
    }
    return;
  }
  int mid = (start + end) / 2;
  update(node * 2, start, mid, l, r, val);
  update(node * 2 + 1, mid + 1, end, l, r, val);
  tree[node] = max(tree[node * 2], tree[node * 2 + 1]);
}
int query(int node, int start, int end, int l, int r) {
  if (start > end || start > r || end < l) return 0;
  if (lazy[node] != 0) {
    tree[node] += lazy[node];
    if (start != end) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start >= l && end <= r) return tree[node];
  int mid = (start + end) / 2;
  int p1 = query(node * 2, start, mid, l, r);
  int p2 = query(node * 2 + 1, mid + 1, end, l, r);
  return max(p1, p2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> K;
  for (int i = 1; i < N + 1; i++) cin >> A[i];
  stack<pair<int, int> > st;
  for (int i = N; i >= 1; i--) {
    while (st.size() && st.top().first <= A[i]) st.pop();
    if (st.size()) {
      graph[i].push_back(st.top().second);
      graph[st.top().second].push_back(i);
    }
    st.push({A[i], i});
  }
  for (int i = N; i >= 1; i--)
    if (!seen[i]) {
      dfs(i, i);
      graph[i].push_back(0);
      graph[0].push_back(i);
    }
  dfs2(0, 0);
  for (int i = 1; i < K + 1; i++) update(1, 1, cnt, in[i], out[i], 1);
  vector<int> res;
  res.push_back(query(1, 1, cnt, 1, cnt));
  for (int i = K + 1; i < N + 1; i++) {
    update(1, 1, cnt, in[i - K], out[i - K], -1);
    update(1, 1, cnt, in[i], out[i], 1);
    res.push_back(query(1, 1, cnt, 1, cnt));
  }
  for (auto &x : res) cout << x << ' ';
  cout << '\n';
  return 0;
}
