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 = 1e6 + 10;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, k, a[N], st[N], tp = 0, tag[N << 2], mx[N << 2];
void mdf(int x, int v) { tag[x] += v, mx[x] += v; }
void pushdown(int x) {
  if (tag[x]) mdf((x << 1), tag[x]), mdf((x << 1 | 1), tag[x]), tag[x] = 0;
}
void modify(int L, int R, int v, int x = 1, int l = 1, int r = n) {
  if (L <= l && r <= R) return mdf(x, v);
  pushdown(x);
  int mid = (l + r) >> 1;
  if (L <= mid) modify(L, R, v, (x << 1), l, mid);
  if (R > mid) modify(L, R, v, (x << 1 | 1), mid + 1, r);
  mx[x] = max(mx[(x << 1)], mx[(x << 1 | 1)]);
}
int query(int L, int R, int x = 1, int l = 1, int r = n) {
  if (L <= l && r <= R) return mx[x];
  pushdown(x);
  int mid = (l + r) >> 1, ret = 0;
  if (L <= mid) ret = max(ret, query(L, R, (x << 1), l, mid));
  if (R > mid) ret = max(ret, query(L, R, (x << 1 | 1), mid + 1, r));
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    a[i] = gi();
    while (tp && a[st[tp]] < a[i]) --tp;
    modify(st[tp] + 1, i, 1);
    st[++tp] = i;
    if (i >= k) cout << query(i - k + 1, i) << ' ';
  }
  return 0;
}
