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 = 1123456;
struct Seg {
  int l, r, val, lzy;
} T[maxn << 2];
void build(int o, int l, int r) {
  T[o].l = l, T[o].r = r;
  if (l != r) {
    int m = l + r >> 1;
    build(o << 1, l, m);
    build(o << 1 | 1, m + 1, r);
  }
}
inline void push_down(int o) {
  T[o << 1].lzy += T[o].lzy;
  T[o << 1 | 1].lzy += T[o].lzy;
  T[o].lzy = 0;
}
inline void push_up(int o) {
  T[o].val =
      max(T[o << 1].val + T[o << 1].lzy, T[o << 1 | 1].val + T[o << 1 | 1].lzy);
}
void add(int o, int l, int r) {
  if (l <= T[o].l && T[o].r <= r) {
    T[o].lzy++;
    return;
  }
  push_down(o);
  int m = T[o].l + T[o].r >> 1;
  if (l <= m) add(o << 1, l, r);
  if (r > m) add(o << 1 | 1, l, r);
  push_up(o);
}
int query(int o, int l, int r) {
  if (l <= T[o].l && T[o].r <= r) return T[o].val + T[o].lzy;
  push_down(o);
  int m = T[o].l + T[o].r >> 1, ret = 0;
  if (l <= m) ret = max(ret, query(o << 1, l, r));
  if (r > m) ret = max(ret, query(o << 1 | 1, l, r));
  push_up(o);
  return ret;
}
deque<pair<int, int> > q;
void add(int pos, int a) {
  int cnt = 1;
  while (!q.empty()) {
    if (q.back().first < a) {
      cnt += q.back().second;
      q.pop_back();
    } else if (q.back().first == a) {
      add(1, pos - cnt + 1, pos);
      q.back().second += cnt;
      return;
    } else
      break;
  }
  q.emplace_back(a, cnt);
  add(1, pos - cnt + 1, pos);
}
void del() {
  if (q.front().second == 1)
    q.pop_front();
  else
    q.front().second--;
}
int k, n;
int a[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int(i) = (1); (i) <= (n); (i)++) scanf("%d", a + i);
  build(1, 1, 1 << 20);
  for (int(i) = (1); (i) <= (k); (i)++) add(i, a[i]);
  printf("%d", query(1, 1, k));
  for (int(i) = (k + 1); (i) <= (n); (i)++) {
    del();
    add(i, a[i]);
    printf(" %d", query(1, i - k + 1, i));
  }
}
