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 long long inff = 0x3f3f3f3f3f3f3f3f;
int n, k, a[1000008], de[1000008], qw, num, ls[1000008], rs[1000008],
    deep[1000008], s[1000008], zz;
bool vis[1000008];
vector<int> g[1000008];
struct as {
  int l, r, ma, lz;
} tr[1000008 << 2];
void push(int d) { tr[d].ma = max(tr[d << 1].ma, tr[d << 1 | 1].ma); }
void build(int d, int l, int r) {
  tr[d].l = l, tr[d].r = r, tr[d].lz = 0;
  if (l == r) {
    tr[d].ma = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(d << 1, l, mid);
  build(d << 1 | 1, mid + 1, r);
  push(d);
}
void pussh(int d) {
  if (!tr[d].lz) return;
  tr[d << 1].lz += tr[d].lz, tr[d << 1 | 1].lz += tr[d].lz;
  tr[d << 1].ma += tr[d].lz, tr[d << 1 | 1].ma += tr[d].lz;
  tr[d].lz = 0;
}
void add(int d, int l, int r, int pos) {
  if (tr[d].l == l && tr[d].r == r) {
    tr[d].ma += pos, tr[d].lz += pos;
    return;
  }
  pussh(d);
  int mid = (tr[d].l + tr[d].r) >> 1;
  if (mid >= r)
    add(d << 1, l, r, pos);
  else if (l > mid)
    add(d << 1 | 1, l, r, pos);
  else
    add(d << 1, l, mid, pos), add(d << 1 | 1, mid + 1, r, pos);
  push(d);
}
int query(int d, int l, int r) {
  if (tr[d].l == l && tr[d].r == r) {
    return tr[d].ma;
  }
  int mid = (tr[d].l + tr[d].r) >> 1;
  if (mid >= r)
    return query(d << 1, l, r);
  else if (l > mid)
    return query(d << 1 | 1, l, r);
  else
    return max(query(d << 1, l, mid), query(d << 1 | 1, mid + 1, r));
}
void dfs(int x) {
  ls[x] = ++num;
  vis[x] = 1;
  for (int i = 0; i < g[x].size(); i++)
    deep[g[x][i]] = deep[x] + 1, dfs(g[x][i]);
  rs[x] = num;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  int l = 1, r = 1;
  for (int i(1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    while (r > l && a[i] > a[de[r - 1]]) g[i].push_back(de[r - 1]), r--;
    de[r++] = i;
  }
  for (int i(n); i >= (1); --i)
    if (!vis[i]) deep[i] = 1, dfs(i);
  build(1, 1, n);
  for (int i(n); i >= (1); --i) {
    add(1, ls[i], ls[i], deep[i]);
    if (i + k - 1 > n) continue;
    int ans = query(1, ls[i + k - 1], ls[i]);
    s[++zz] = ans;
    add(1, ls[i + k - 1], ls[i + k - 1], -int(0x3f3f3f3f));
    if (ls[i + k - 1] != rs[i + k - 1])
      add(1, ls[i + k - 1] + 1, rs[i + k - 1], -1);
  }
  for (int i(zz); i >= (1); --i) printf("%d ", s[i]);
  puts("");
  return 0;
}
