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 ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
const int MN = 1e6 + 5;
int n, k, a[MN];
vi g[MN];
int tin[MN], tout[MN], tt;
template <typename T>
struct SegmentTree {
  T t[MN * 4], lz[MN * 4];
  const T ID = 0;
  T merge(T x, T y) { return max(x, y); }
  void push(int i, int l, int r) {
    if (lz[i]) {
      t[i] += lz[i];
      if (l != r) {
        lz[(i << 1)] += lz[i];
        lz[(i << 1 | 1)] += lz[i];
      }
      lz[i] = 0;
    }
  }
  void upd(int ql, int qr, int v, int i = 1, int l = 1, int r = n + 1) {
    push(i, l, r);
    if (l > qr || r < ql || ql > qr) return;
    if (l >= ql && r <= qr) {
      lz[i] += v, push(i, l, r);
      return;
    }
    int m = (l + r) / 2;
    upd(ql, qr, v, (i << 1), l, m), upd(ql, qr, v, (i << 1 | 1), m + 1, r);
    t[i] = merge(t[(i << 1)], t[(i << 1 | 1)]);
  }
  T get(int ql, int qr, int i = 1, int l = 1, int r = n + 1) {
    push(i, l, r);
    if (l > qr || r < ql) return ID;
    if (l >= ql && r <= qr) return t[i];
    int m = (l + r) / 2;
    return merge(get(ql, qr, (i << 1), l, m),
                 get(ql, qr, (i << 1 | 1), m + 1, r));
  }
};
SegmentTree<int> seg;
void ett(int u) {
  tin[u] = ++tt;
  for (int v : g[u]) ett(v);
  tout[u] = tt;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  stack<int> s;
  for (int i = n - 1; i >= 0; i--) {
    while (!s.empty() && a[i] >= a[s.top()]) s.pop();
    int nxt = s.empty() ? n : s.top();
    g[nxt].push_back(i);
    s.push(i);
  }
  ett(n);
  for (int i = 0; i < k - 1; i++) seg.upd(tin[i], tout[i], 1);
  for (int i = k - 1; i < n; i++) {
    seg.upd(tin[i], tout[i], 1);
    cout << seg.get(1, n + 1) << " ";
    seg.upd(tin[i - k + 1], tout[i - k + 1], -1);
  }
  return 0;
}
