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;
const int sz = 1e6 + 527;
int n, k;
int cnt, T;
int x, y, z;
int head[sz];
int a[sz], ans[sz];
int dfn[sz], lev[sz];
stack<int> s;
struct Edge {
  int v, nxt;
} edge[sz];
struct node {
  int tag, mx;
} tr[sz << 2];
void add(int u, int v) {
  edge[++cnt] = (Edge){v, head[u]};
  head[u] = cnt;
}
void update(int o) {
  tr[o << 1].tag += tr[o].tag;
  tr[o << 1 | 1].tag += tr[o].tag;
  tr[o << 1].mx += tr[o].tag;
  tr[o << 1 | 1].mx += tr[o].tag;
  tr[o].tag = 0;
}
void modify(int o, int l, int r) {
  if (x <= l && r <= y) return (void)(tr[o].mx += z, tr[o].tag += z);
  if (tr[o].tag) update(o);
  int mid = (l + r) >> 1;
  if (x <= mid) modify(o << 1, l, mid);
  if (y > mid) modify(o << 1 | 1, mid + 1, r);
  tr[o].mx = max(tr[o << 1].mx, tr[o << 1 | 1].mx);
}
void dfs(int x) {
  dfn[x] = ++T;
  for (int i = head[x]; i; i = edge[i].nxt) dfs(edge[i].v);
  lev[x] = T;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    while (!s.empty() && a[s.top()] < a[i]) {
      add(i, s.top());
      s.pop();
    }
    s.push(i);
  }
  n++;
  while (!s.empty()) {
    add(n, s.top());
    s.pop();
  }
  dfs(n);
  for (int i = 1; i <= k; i++) {
    x = dfn[i], y = lev[i], z = 1;
    modify(1, 1, n);
  }
  ans[1] = tr[1].mx;
  for (int i = k + 1; i < n; i++) {
    x = dfn[i], y = lev[i], z = 1;
    modify(1, 1, n);
    x = dfn[i - k], y = lev[i - k], z = -1;
    modify(1, 1, n);
    ans[i - k + 1] = tr[1].mx;
  }
  for (int i = 1; i <= n - k; i++) printf("%d ", ans[i]);
}
