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 + 6, LG = 22;
int n, k;
int arr[N], nxt[N];
vector<int> adj[N];
int st[N], en[N], eul[N], dfstime;
int btree[4 * N], lazy[4 * N];
void dfs(int x) {
  st[x] = dfstime;
  eul[dfstime++] = x;
  for (int i : adj[x]) dfs(i);
  en[x] = dfstime - 1;
}
void prob(int node, int l, int r) {
  btree[node] += lazy[node];
  if (l != r) {
    lazy[node << 1] += lazy[node];
    lazy[(node << 1) | 1] += lazy[node];
  }
  lazy[node] = 0;
}
void update(int node, int l, int r, int ql, int qr, int val) {
  if (lazy[node]) prob(node, l, r);
  if (r < ql || qr < l) return;
  if (ql <= l && r <= qr) {
    lazy[node] += val;
    prob(node, l, r);
    return;
  }
  int mid = (l + r) >> 1;
  update(node << 1, l, mid, ql, qr, val);
  update((node << 1) | 1, mid + 1, r, ql, qr, val);
  btree[node] = max(btree[node << 1], btree[(node << 1) | 1]);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  stack<int> stk;
  for (int i = n - 1; i >= 0; i--) {
    while (stk.size() && arr[stk.top()] <= arr[i]) stk.pop();
    if (stk.size())
      nxt[i] = stk.top();
    else
      nxt[i] = n;
    adj[nxt[i]].push_back(i);
    stk.push(i);
  }
  dfs(n);
  for (int i = 0; i < k; i++) {
    update(1, 0, n, st[i], en[i], 1);
  }
  printf("%d", btree[1]);
  for (int i = k; i < n; i++) {
    update(1, 0, n, st[i - k], en[i - k], -1);
    update(1, 0, n, st[i], en[i], 1);
    prob(1, 0, n);
    printf(" %d", btree[1]);
  }
  puts("");
}
