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;
const int maxnode = maxn << 2;
int n, kk, ai[maxn + 5], st[maxn + 5];
int in[maxn + 5], out[maxn + 5], DFN = 0;
int ans[maxn + 5], m = 0;
vector<int> G[maxn + 5];
int tr[maxnode + 5], lz[maxnode + 5];
void PushDown(int x) {
  tr[(x << 1)] += lz[x], tr[(x << 1 | 1)] += lz[x];
  lz[(x << 1)] += lz[x], lz[(x << 1 | 1)] += lz[x];
  lz[x] = 0;
}
void Inc(int x, int l, int r, int ll, int rr) {
  if (ll <= l && r <= rr)
    ++tr[x], ++lz[x];
  else {
    PushDown(x);
    if (ll <= ((l + r) >> 1)) Inc((x << 1), l, ((l + r) >> 1), ll, rr);
    if (((l + r) >> 1) < rr) Inc((x << 1 | 1), ((l + r) >> 1) + 1, r, ll, rr);
    tr[x] = max(tr[(x << 1)], tr[(x << 1 | 1)]);
  }
}
void Reset(int x, int l, int r, int p) {
  if (l == r)
    lz[x] = 0, tr[x] = -(maxn + 10);
  else {
    PushDown(x);
    if (p <= ((l + r) >> 1))
      Reset((x << 1), l, ((l + r) >> 1), p);
    else
      Reset((x << 1 | 1), ((l + r) >> 1) + 1, r, p);
    tr[x] = max(tr[(x << 1)], tr[(x << 1 | 1)]);
  }
}
void dfs(int x) {
  in[x] = ++DFN;
  for (auto y : G[x]) dfs(y);
  out[x] = DFN;
}
int main() {
  scanf("%d%d", &n, &kk);
  for (int i = 1; i <= n; ++i) scanf("%d", ai + i);
  {
    int top = 0;
    for (int i = n; i >= 1; --i) {
      while (top && ai[i] >= ai[st[top]]) --top;
      if (top == 0)
        G[n + 1].push_back(i);
      else
        G[st[top]].push_back(i);
      st[++top] = i;
    }
  }
  dfs(n + 1);
  for (int i = 1; i <= n; ++i) {
    Inc(1, 1, n + 1, in[i], out[i]);
    if (i - kk >= 1) Reset(1, 1, n + 1, in[i - kk]);
    if (i >= kk) ans[++m] = tr[1];
  }
  for (int i = 1; i <= m; ++i) printf("%d%c", ans[i], (i == m) ? '\n' : ' ');
  return 0;
}
