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>
const int N = 1e6 + 10;
using namespace std;
int n, m;
vector<int> g[N];
int a[N], b[N], l[N], r[N], fa[N];
struct node {
  int sum;
  int mx, tag;
} f[N * 4];
void down(int l, int r, int s) {
  if (f[s].tag) {
    if (f[s].sum) f[s].mx += f[s].tag;
    if (l != r) f[s << 1].tag += f[s].tag, f[s << 1 | 1].tag += f[s].tag;
    f[s].tag = 0;
  }
}
void combine(int s) {
  f[s].sum = f[s + s].sum + f[s + s + 1].sum;
  f[s].mx = max(f[s + s].mx, f[s + s + 1].mx);
}
void ins(int l, int r, int s, int ll, bool sig) {
  down(l, r, s);
  if (r < ll || l > ll) return;
  if (l == r) {
    if (sig)
      f[s].sum = 1, f[s].mx = 0;
    else
      f[s].sum = f[s].mx = 0;
    return;
  }
  ins(l, (l + r) / 2, s + s, ll, sig);
  ins((l + r) / 2 + 1, r, s + s + 1, ll, sig);
  combine(s);
}
void add(int l, int r, int s, int ll, int rr, int tag) {
  down(l, r, s);
  if (r < ll || rr < l) return;
  if (ll <= l && r <= rr) {
    f[s].tag += tag;
    down(l, r, s);
    return;
  }
  add(l, (l + r) / 2, s << 1, ll, rr, tag);
  add((l + r) / 2 + 1, r, s + s + 1, ll, rr, tag);
  combine(s);
}
void dfs(int x) {
  static int tot = 0;
  l[x] = ++tot;
  for (auto u : g[x])
    if (u != fa[x]) {
      fa[u] = x;
      dfs(u);
    }
  r[x] = tot;
}
void ins(int x, int y) {
  while (x) {
    b[x] = y;
    x -= x & -x;
  }
}
int get(int x) {
  int v = 1e9;
  while (x <= n) {
    v = min(v, b[x]);
    x += x & -x;
  }
  return v;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) b[i] = 1e9;
  for (int i = n; i >= 1; i--) {
    int v = get(a[i]);
    if (v != 1e9) g[v].push_back(i);
    ins(a[i] - 1, i);
  }
  for (int i = n; i >= 1; i--)
    if (!l[i]) dfs(i);
  for (int i = 1; i <= m; i++) {
    add(1, n, 1, l[i], r[i], 1);
    ins(1, n, 1, l[i], 1);
  }
  for (int i = 1; i + m - 1 <= n; i++) {
    printf("%d ", f[1].mx + 1);
    ins(1, n, 1, l[i], 0);
    add(1, n, 1, l[i + m], r[i + m], 1);
    ins(1, n, 1, l[i + m], 1);
  }
  return 0;
}
