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;
using ii = pair<int, int>;
const int MAX = 1e6 + 6;
int n, k, a[MAX];
int p[MAX];
int compnxt(int i) {
  if (p[i]) return p[i];
  p[i] = i + 1;
  while (p[i] <= n && a[p[i]] <= a[i]) {
    p[i] = compnxt(p[i]);
  }
  return p[i];
}
vector<int> ch[MAX];
int disc[MAX], leave[MAX], dfstack[MAX];
void dfs() {
  int dfst = 0;
  int s = 0;
  dfstack[s++] = n + 1;
  while (s) {
    int u = dfstack[s - 1];
    if (disc[u]) {
      leave[u] = ++dfst;
      --s;
    } else {
      disc[u] = ++dfst;
      for (int v : ch[u]) {
        dfstack[s++] = v;
      }
    }
  }
  assert(dfst == 2 * (n + 1));
}
int seg[8 * MAX], lazy[8 * MAX];
void pushdown(int u) {
  lazy[2 * u] += lazy[u];
  seg[2 * u] += lazy[u];
  lazy[2 * u + 1] += lazy[u];
  seg[2 * u + 1] += lazy[u];
  lazy[u] = 0;
}
void upd(int u, int tl, int tr, int l, int r, int v) {
  if (l > tr || r < tl) return;
  if (lazy[u] && tl != tr) pushdown(u);
  if (l <= tl && tr <= r) {
    seg[u] += v;
    lazy[u] = v;
  } else {
    int tm = (tl + tr) / 2;
    upd(2 * u, tl, tm, l, r, v);
    upd(2 * u + 1, tm + 1, tr, l, r, v);
    seg[u] = max(seg[2 * u], seg[2 * u + 1]);
  }
}
int ans[MAX];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) {
    compnxt(i);
    ch[p[i]].push_back(i);
  }
  dfs();
  for (int i = n; i; --i) {
    if (i + k <= n) {
      upd(1, 1, 2 * (n + 1), disc[i + k], leave[i + k], -1);
    }
    upd(1, 1, 2 * (n + 1), disc[i], leave[i], 1);
    if (i <= n - k + 1) {
      ans[i] = seg[1];
    }
  }
  for (int i = 1; i <= n - k + 1; ++i) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}
