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;
int n, k;
int arr[1000008];
int l[1000008];
int mx[1000008 << 2], lazy[1000008 << 2];
void pushup(int rt) { mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]); }
void build(int rt, int l, int r) {
  if (l > r) return;
  if (l == r) {
    mx[l] = arr[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  pushup(rt);
}
void pushdown(int rt) {
  mx[rt << 1] += lazy[rt];
  mx[rt << 1 | 1] += lazy[rt];
  lazy[rt << 1] += lazy[rt];
  lazy[rt << 1 | 1] += lazy[rt];
  lazy[rt] = 0;
}
void update(int rt, int l, int r, int L, int R, int C) {
  if (L <= l && r <= R) {
    lazy[rt] += C;
    mx[rt] += C;
    return;
  }
  pushdown(rt);
  int mid = (l + r) >> 1;
  if (L <= mid) update(rt << 1, l, mid, L, R, C);
  if (mid < R) update(rt << 1 | 1, mid + 1, r, L, R, C);
  pushup(rt);
}
int query(int rt, int l, int r, int L, int R) {
  if (L <= l && r <= R) return mx[rt];
  int mid = (l + r) >> 1;
  pushdown(rt);
  int ret = -1;
  if (L <= mid) ret = max(ret, query(rt << 1, l, mid, L, R));
  if (mid < R) ret = max(ret, query(rt << 1 | 1, mid + 1, r, L, R));
  return ret;
}
stack<int> sta;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
  for (int i = n; i >= 1; --i) {
    while (!sta.empty() && arr[sta.top()] <= arr[i]) {
      l[sta.top()] = i + 1;
      sta.pop();
    }
    sta.push(i);
  }
  while (!sta.empty()) {
    l[sta.top()] = 1;
    sta.pop();
  }
  for (int i = 1; i <= n; ++i) {
    update(1, 1, n, l[i], i, 1);
    if (i >= k) {
      printf("%d ", query(1, 1, n, i - k + 1, i));
      update(1, 1, n, l[i - k + 1], i - k + 1, -1);
    }
  }
  return 0;
}
