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 vals[1123456];
vector<int> adj[1123456];
int timestamp;
int Tin[1123456], Tout[1123456];
int stree[4 * 1123456];
int lazy_tree[4 * 1123456];
void ETT(int u) {
  Tin[u] = timestamp;
  for (int v : adj[u]) ETT(v);
  Tout[u] = timestamp++;
}
void generate_tree() {
  stack<pair<int, int> > gt_stack;
  for (int i = N; i >= 1; i--) {
    while (!gt_stack.empty() && gt_stack.top().first <= vals[i]) {
      gt_stack.pop();
    }
    if (gt_stack.empty()) {
      adj[0].push_back(i);
    } else {
      adj[gt_stack.top().second].push_back(i);
    }
    gt_stack.push({vals[i], i});
  }
  ETT(0);
  ;
}
void lazy_update(int node, int l, int r) {
  stree[node] += lazy_tree[node];
  if (l != r) {
    lazy_tree[node + node + 1] += lazy_tree[node];
    lazy_tree[node + node + 2] += lazy_tree[node];
  }
  lazy_tree[node] = 0;
}
void update(int node, int l, int r, int tl, int tr, int offset) {
  lazy_update(node, l, r);
  if (l > tr || r < tl) {
    return;
  }
  if (l >= tl && r <= tr) {
    lazy_tree[node] += offset;
    lazy_update(node, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(node + node + 1, l, mid, tl, tr, offset);
  update(node + node + 2, mid + 1, r, tl, tr, offset);
  stree[node] = max(stree[node + node + 1], stree[node + node + 2]);
}
int query(int node, int l, int r, int tl, int tr) {
  lazy_update(node, l, r);
  if (l > tr || r < tl) {
    return 0;
  }
  if (l >= tl && r <= tr) {
    return stree[node];
  }
  int mid = (l + r) / 2;
  return query(node + node + 1, l, mid, tl, tr) +
         query(node + node + 2, mid + 1, r, tl, tr);
}
void enable(int u) { update(0, 0, N, Tin[u], Tout[u], 1); }
void disable(int u) { update(0, 0, N, Tin[u], Tout[u], -1); }
void print_seg() {
  for (int i = 0; i < N; i++) {
    printf("%d ", query(0, 0, N, i, i));
  }
  printf("\n");
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &vals[i]);
  }
  generate_tree();
  for (int i = 1; i <= N; i++) {
    enable(i);
    if (i >= K) {
      printf("%d\n", query(0, 0, N - 1, 0, N - 1));
      disable(i - K + 1);
    }
  }
}
