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 + 10;
int n, k, p[maxn], stk[maxn], top, head[maxn], tid[maxn], siz[maxn], tot, cnt;
struct abc {
  int to, nxt;
} e[maxn * 2];
void add(int x, int y) {
  e[++cnt].nxt = head[x];
  e[cnt].to = y;
  head[x] = cnt;
}
void dfs(int x) {
  tid[x] = ++tot;
  siz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    dfs(e[i].to);
    siz[x] += siz[e[i].to];
  }
}
int tree[4 * maxn], tag[4 * maxn];
void down(int o) {
  tree[o << 1] += tag[o];
  tree[o << 1 | 1] += tag[o];
  tag[o << 1] += tag[o];
  tag[o << 1 | 1] += tag[o];
  tag[o] = 0;
}
void update(int l, int r, int o, int x, int y, int val) {
  if (x <= l && r <= y) {
    tree[o] += val;
    tag[o] += val;
    return;
  }
  int mid = (l + r) >> 1;
  down(o);
  if (x <= mid) update(l, mid, o << 1, x, y, val);
  if (y > mid) update(mid + 1, r, o << 1 | 1, x, y, val);
  tree[o] = max(tree[o << 1], tree[o << 1 | 1]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 1; i <= n; ++i) {
    while (top && p[stk[top]] < p[i]) {
      add(i, stk[top]);
      --top;
    }
    stk[++top] = i;
  }
  while (top) {
    add(n + 1, stk[top]);
    --top;
  }
  dfs(n + 1);
  for (int i = 1; i <= k; ++i)
    update(1, n + 1, 1, tid[i], tid[i] + siz[i] - 1, 1);
  printf("%d", tree[1]);
  for (int i = k + 1; i <= n; ++i) {
    update(1, n + 1, 1, tid[i], tid[i] + siz[i] - 1, 1);
    update(1, n + 1, 1, tid[i - k], tid[i - k] + siz[i - k] - 1, -1);
    printf(" %d", tree[1]);
  }
}
