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, m, m1, len, top, a[1000005], size[1000005], dfn[1000005], nex[1000005],
    wen[1000005], hea[1000005], sta[1000005];
struct segment_tree {
  int a[4000005], lazy[4000005];
  void pushdown(int k) {
    int t = lazy[k];
    a[k << 1] += t;
    lazy[k << 1] += t;
    a[(k << 1) | 1] += t;
    lazy[(k << 1) | 1] += t;
    lazy[k] = 0;
  }
  void update(int l, int r, int k, int x, int y, int z) {
    if (l >= x && r <= y) {
      a[k] += z;
      lazy[k] += z;
      return;
    }
    pushdown(k);
    int mid = (l + r) >> 1;
    if (x <= mid) update(l, mid, k << 1, x, y, z);
    if (y > mid) update(mid + 1, r, (k << 1) | 1, x, y, z);
    a[k] = max(a[k << 1], a[(k << 1) | 1]);
  }
  int find(int l, int r, int k, int x, int y) {
    if (l >= x && r <= y) return a[k];
    pushdown(k);
    int mid = (l + r) >> 1, ans = 0;
    if (x <= mid) ans = find(l, mid, k << 1, x, y);
    if (y > mid) ans = max(ans, find(mid + 1, r, (k << 1) | 1, x, y));
    return ans;
  }
} st;
void add(int x, int y) {
  ++len;
  nex[len] = hea[x];
  wen[len] = y;
  hea[x] = len;
}
void dfs(int x) {
  dfn[x] = ++m1;
  size[x] = 1;
  for (int i = hea[x]; i; i = nex[i]) {
    dfs(wen[i]);
    size[x] += size[wen[i]];
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[n + 1] = n + 1;
  ++n;
  for (int i = 1; i <= n; i++) {
    while (top && a[sta[top]] < a[i]) add(i, sta[top--]);
    sta[++top] = i;
  }
  dfs(n);
  for (int i = 1; i <= m; i++)
    st.update(1, n, 1, dfn[i], dfn[i] + size[i] - 1, 1);
  printf("%d", st.a[1]);
  for (int i = m + 1; i < n; i++) {
    st.update(1, n, 1, dfn[i], dfn[i] + size[i] - 1, 1);
    st.update(1, n, 1, dfn[i - m], dfn[i - m] + size[i - m] - 1, -1);
    printf(" %d", st.a[1]);
  }
}
