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 + 100;
struct segtree {
  int mx[maxn << 2], lazy[maxn << 2];
  void maintain(int rt) { mx[rt] = max(mx[(rt << 1)], mx[(rt << 1 | 1)]); }
  void push(int rt) {
    if (!lazy[rt]) return;
    lazy[(rt << 1)] += lazy[rt], lazy[(rt << 1 | 1)] += lazy[rt];
    mx[(rt << 1)] += lazy[rt], mx[(rt << 1 | 1)] += lazy[rt];
    lazy[rt] = 0;
  }
  void update(int l, int r, int rt, int ql, int qr, int v) {
    if (ql <= l && qr >= r) {
      lazy[rt] += v;
      mx[rt] += v;
      return;
      ;
    }
    push(rt);
    int mid = l + r >> 1;
    if (ql <= mid) update(l, mid, (rt << 1), ql, qr, v);
    if (qr > mid) update(mid + 1, r, (rt << 1 | 1), ql, qr, v);
    maintain(rt);
  }
} T;
vector<vector<int> > v;
int in[maxn], out[maxn], tot;
void dfs(int x) {
  in[x] = ++tot;
  for (auto &u : v[x]) dfs(u);
  out[x] = tot;
}
int go[maxn], a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  v.resize(n + 2);
  for (int i = 1; i <= n; ++i) cin >> a[i], go[i] = i + 1;
  for (int i = n; i >= 1; --i) {
    while (go[i] <= n && a[i] >= a[go[i]]) {
      go[i] = go[go[i]];
    }
    v[go[i]].push_back(i);
  }
  dfs(n + 1);
  for (int i = 1; i <= n; ++i) {
    T.update(1, n + 1, 1, in[i], out[i], 1);
    if (i - k >= 1)
      T.update(1, n + 1, 1, in[i - k], out[i - k], -1),
          T.update(1, n + 1, 1, in[i - k], in[i - k], -1e9);
    if (i - k >= 0) cout << T.mx[1] << ' ';
  }
  cout << '\n';
  return 0;
}
