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 N = 1000010;
inline int read() {
  int s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
int n, K, a[N], pos[N];
vector<pair<int, int> > A;
set<int> B;
int sum[N << 2], flag[N << 2];
inline void pu(int x) { sum[x] = max(sum[(x << 1)], sum[(x << 1 | 1)]); }
inline void pd(int x) {
  sum[(x << 1)] += flag[x];
  sum[(x << 1 | 1)] += flag[x];
  flag[(x << 1)] += flag[x];
  flag[(x << 1 | 1)] += flag[x];
  flag[x] = 0;
}
void UpDate(int u, int v, int l, int r, int x, int k) {
  if (l >= u && r <= v) {
    sum[x] += k;
    flag[x] += k;
    return;
  }
  if (flag[x]) pd(x);
  int mid = (l + r) / 2;
  if (u <= mid) UpDate(u, v, l, mid, (x << 1), k);
  if (v > mid) UpDate(u, v, mid + 1, r, (x << 1 | 1), k);
  pu(x);
}
int Ask(int u, int v, int l, int r, int x) {
  if (l >= u && r <= v) return sum[x];
  if (flag[x]) pd(x);
  int mid = (l + r) / 2, tt = 0;
  if (u <= mid) tt = max(tt, Ask(u, v, l, mid, (x << 1)));
  if (v > mid) tt = max(tt, Ask(u, v, mid + 1, r, (x << 1 | 1)));
  return tt;
}
signed main() {
  n = read(), K = read();
  for (register int i = 1; i <= n; i++)
    a[i] = read(), A.emplace_back(make_pair(a[i], -i));
  sort(A.begin(), A.end());
  reverse(A.begin(), A.end());
  B.insert(0), B.insert(n + 1);
  for (register int i = 0; i < n; i++) {
    int id = -A[i].second;
    int tt = *(--B.lower_bound(id));
    B.insert(id), pos[id] = tt + 1;
  }
  for (register int i = 1; i <= n; i++) {
    UpDate(pos[i], i, 1, n, 1, 1);
    if (i >= K) printf("%d ", Ask(i - K + 1, i, 1, n, 1));
  }
  puts("");
  return 0;
}
