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;
vector<int> v[1000005];
int ans[1000005];
struct node {
  node *l, *r;
  int a, b;
  long long Max;
  long long tag;
  node(int _a, int _b) : l(NULL), r(NULL), a(_a), b(_b), Max(0), tag(0) {}
} * root;
long long Max(node *n) { return n->Max + n->tag; }
void push(node *n) {
  n->l->tag += n->tag;
  n->r->tag += n->tag;
  n->tag = 0;
}
void pull(node *n) { n->Max = max(Max(n->l), Max(n->r)); }
void build(node *n) {
  if (n->a == n->b) return;
  int mid = (n->a + n->b) / 2;
  n->l = new node(n->a, mid);
  n->r = new node(mid + 1, n->b);
  build(n->l);
  build(n->r);
}
void add(node *n, int l, int r, int k) {
  if (n->a >= l && n->b <= r) {
    n->tag += k;
    return;
  }
  if (n->b < l || n->a > r) return;
  push(n);
  add(n->l, l, r, k);
  add(n->r, l, r, k);
  pull(n);
}
int in[1000005];
int out[1000005];
int t;
int a[1000005];
void dfs(int first) {
  in[first] = ++t;
  for (auto it : v[first]) {
    dfs(it);
  }
  out[first] = t;
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[n + 1] = 1e9;
  vector<pair<int, int> > mystk;
  mystk.push_back(make_pair(1e9 + 1, 0));
  for (int i = 1; i <= n + 1; i++) {
    while (mystk.back().first < a[i]) {
      v[i].push_back(mystk.back().second);
      mystk.pop_back();
    }
    mystk.push_back(make_pair(a[i], i));
  }
  dfs(n + 1);
  root = new node(1, n + 1);
  build(root);
  for (int i = 1; i <= k; i++) {
    add(root, in[i], out[i], 1);
  }
  for (int i = k + 1; i <= n + 1; i++) {
    printf("%lld ", Max(root));
    add(root, in[i - k], in[i - k], -1e9);
    add(root, in[i], out[i], 1);
  }
}
