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 MAXN = 1e6 + 10;
const double eps = 1e-8;
using namespace std;
const int inf = 1e9;
struct edge {
  int t;
  edge *next;
} e[MAXN << 1], *h[MAXN], *o = e;
void add(int x, int y) {
  o->t = y;
  o->next = h[x];
  h[x] = o++;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int n, k;
int a[MAXN], key[MAXN];
int st[MAXN], tot, cnt;
int fa[MAXN][21], dep[MAXN], p[MAXN], num[MAXN];
void dfs(int x, int pre, int deep) {
  fa[x][0] = pre;
  dep[x] = deep + 1;
  num[x] = 1;
  p[x] = ++cnt;
  key[x] = x;
  for (int i = 1; i <= 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (edge *j = h[x]; j; j = j->next) {
    dfs(j->t, x, deep + 1);
    num[x] += num[j->t];
  }
}
int St(int x, int y) {
  int p = x;
  for (int i = 20; i >= 0; i--)
    if (key[fa[x][i]] <= y) x = fa[x][i];
  return dep[p] - dep[fa[x][0]];
}
int maxx[MAXN << 2], tag[MAXN << 2];
void push(int x) {
  if (tag[x]) {
    maxx[x << 1] -= tag[x];
    maxx[x << 1 | 1] -= tag[x];
    tag[x << 1] += tag[x];
    tag[x << 1 | 1] += tag[x];
    tag[x] = 0;
  }
}
void up(int x) { maxx[x] = max(maxx[x << 1], maxx[x << 1 | 1]); }
void built(int x, int l, int r) {
  if (l == r) {
    maxx[x] = -inf;
    return;
  }
  int mid = (l + r) >> 1;
  built(x << 1, l, mid);
  built(x << 1 | 1, mid + 1, r);
  up(x);
}
void update(int x, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) {
    tag[x]++;
    maxx[x]--;
    return;
  }
  int mid = (l + r) >> 1;
  push(x);
  if (ql <= mid) update(x << 1, l, mid, ql, qr);
  if (qr > mid) update(x << 1 | 1, mid + 1, r, ql, qr);
  up(x);
}
void update1(int x, int l, int r, int t, int k) {
  if (l == r) {
    maxx[x] += k;
    return;
  }
  int mid = (l + r) >> 1;
  push(x);
  if (t <= mid)
    update1(x << 1, l, mid, t, k);
  else
    update1(x << 1 | 1, mid + 1, r, t, k);
  up(x);
}
int ans[MAXN];
int main() {
  n = read();
  k = read();
  key[0] = inf;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = n; i >= 1; i--) {
    while (tot && a[st[tot]] <= a[i]) tot--;
    if (!tot)
      add(n + 1, i);
    else
      add(st[tot], i);
    st[++tot] = i;
  }
  dfs(n + 1, 0, 0);
  built(1, 1, n + 1);
  for (int i = n; i >= n - k + 1; i--)
    update1(1, 1, n + 1, p[i], dep[i] - 1 + inf);
  ans[n - k + 1] = maxx[1];
  for (int i = n - k; i >= 1; i--) {
    update(1, 1, n + 1, p[i + k], p[i + k] + num[i + k] - 1);
    update1(1, 1, n + 1, p[i + k], -inf);
    update1(1, 1, n + 1, p[i], dep[i] + inf - 1);
    ans[i] = maxx[1];
  }
  for (int i = 1; i <= n - k + 1; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}
