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 a[1000005];
struct edge {
  int from;
  int to;
  int next;
} E[1000005 << 1];
int head[1000005];
int sz = 1;
void add_edge(int u, int v) {
  sz++;
  E[sz].from = u;
  E[sz].to = v;
  E[sz].next = head[u];
  head[u] = sz;
}
int cnt = 0;
int lb[1000005], rb[1000005];
void dfs(int x, int fa) {
  lb[x] = ++cnt;
  for (int i = head[x]; i; i = E[i].next) {
    int y = E[i].to;
    if (y != fa) {
      dfs(y, x);
    }
  }
  rb[x] = cnt;
}
struct node {
  int l;
  int r;
  int v;
  int mark;
} tree[1000005 << 2];
void push_up(int pos) {
  tree[pos].v = max(tree[pos << 1].v, tree[pos << 1 | 1].v);
}
void build(int l, int r, int pos) {
  tree[pos].l = l;
  tree[pos].r = r;
  if (l == r) {
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, pos << 1);
  build(mid + 1, r, pos << 1 | 1);
  push_up(pos);
}
void push_down(int pos) {
  if (tree[pos].mark) {
    tree[pos << 1].v += tree[pos].mark;
    tree[pos << 1].mark += tree[pos].mark;
    tree[pos << 1 | 1].v += tree[pos].mark;
    tree[pos << 1 | 1].mark += tree[pos].mark;
    tree[pos].mark = 0;
  }
}
void update(int L, int R, int v, int pos) {
  if (L <= tree[pos].l && R >= tree[pos].r) {
    tree[pos].v += v;
    tree[pos].mark += v;
    return;
  }
  push_down(pos);
  int mid = (tree[pos].l + tree[pos].r) >> 1;
  if (L <= mid) update(L, R, v, pos << 1);
  if (R > mid) update(L, R, v, pos << 1 | 1);
  push_up(pos);
}
int query(int L, int R, int pos) {
  if (L <= tree[pos].l && R >= tree[pos].r) {
    return tree[pos].v;
  }
  push_down(pos);
  int mid = (tree[pos].l + tree[pos].r) >> 1;
  int ans = 0;
  if (L <= mid) ans = max(ans, query(L, R, pos << 1));
  if (R > mid) ans = max(ans, query(L, R, pos << 1 | 1));
  return ans;
}
int nex[1000005];
void init() {
  stack<int> s;
  for (int i = n; i >= 1; i--) {
    while (!s.empty() && a[s.top()] <= a[i]) s.pop();
    if (!s.empty()) {
      int p = s.top();
      add_edge(p, i);
      add_edge(i, p);
    } else {
      add_edge(n + 1, i);
      add_edge(i, n + 1);
    }
    s.push(i);
  }
  dfs(n + 1, 0);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  init();
  build(1, n + 1, 1);
  for (int i = 1; i <= k; i++) {
    update(lb[i], rb[i], 1, 1);
  }
  for (int i = 1; i + k - 1 <= n; i++) {
    int r = i + k - 1;
    printf("%d ", query(1, n + 1, 1));
    update(lb[i], rb[i], -1, 1);
    update(lb[r + 1], rb[r + 1], 1, 1);
  }
}
