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 maxn = 1e6 + 5;
int mx[maxn * 4], tag[maxn * 4], cnt, sz[maxn], a[maxn], id[maxn];
vector<int> G[maxn];
stack<int> s;
void dfs(int u) {
  sz[u] = 1;
  id[u] = ++cnt;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    dfs(v);
    sz[u] += sz[v];
  }
}
void pushdown(int o, int ls, int rs) {
  if (tag[o] != 0) {
    tag[ls] += tag[o], tag[rs] += tag[o];
    mx[ls] += tag[o], mx[rs] += tag[o];
    tag[o] = 0;
  }
}
void up(int o, int l, int r, int ql, int qr, int v) {
  int m = (l + r) / 2, ls = o * 2, rs = o * 2 + 1;
  if (l != r) pushdown(o, ls, rs);
  if (l >= ql && r <= qr) {
    mx[o] += v;
    tag[o] += v;
    return;
  }
  if (ql <= m) up(ls, l, m, ql, qr, v);
  if (qr > m) up(rs, m + 1, r, ql, qr, v);
  mx[o] = max(mx[ls], mx[rs]);
}
int main() {
  int n, k, ans = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = n; i; i--) {
    while (!s.empty() && a[i] >= a[s.top()]) s.pop();
    if (!s.empty()) G[s.top()].push_back(i);
    s.push(i);
  }
  for (int i = n; i; i--)
    if (!id[i]) dfs(i);
  for (int i = 1; i <= k; i++) up(1, 1, n, id[i], id[i] + sz[i] - 1, 1);
  printf("%d", mx[1]);
  for (int i = k + 1; i <= n; i++) {
    up(1, 1, n, id[i - k], id[i - k], -10000000);
    up(1, 1, n, id[i], id[i] + sz[i] - 1, 1);
    printf(" %d", mx[1]);
  }
}
