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 arr[1000005], st[1000005], et[1000005];
int tree[1000005 * 4], lazy[1000005 * 4];
vector<int> vv[1000005], ans;
stack<int> SS;
int tt;
void dfs(int u, int par) {
  st[u] = ++tt;
  for (int i = 0; i < vv[u].size(); i++) {
    int v = vv[u][i];
    if (v != par) dfs(v, u);
  }
  et[u] = tt;
}
void Down(int node, int left, int right) {
  if (lazy[node]) {
    tree[left] += lazy[node];
    tree[right] += lazy[node];
    lazy[left] += lazy[node];
    lazy[right] += lazy[node];
    lazy[node] = 0;
  }
}
void update(int node, int b, int e, int l, int r, int v) {
  if (b > r || e < l) return;
  if (b >= l && e <= r) {
    tree[node] += v;
    lazy[node] += v;
    return;
  }
  int left = 2 * node, right = left + 1, mid = (b + e) / 2;
  Down(node, left, right);
  update(left, b, mid, l, r, v);
  update(right, mid + 1, e, l, r, v);
  tree[node] = max(tree[left], tree[right]);
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  arr[0] = n + 5;
  SS.push(0);
  for (int i = n; i >= 1; i--) {
    while (arr[SS.top()] <= arr[i]) SS.pop();
    if (SS.top() != 0) vv[SS.top()].push_back(i);
    SS.push(i);
  }
  for (int i = n; i >= 1; i--) {
    if (st[i] == 0) dfs(i, 0);
  }
  for (int i = k; i >= 1; i--) {
    update(1, 1, n, st[i], et[i], 1);
  }
  ans.push_back(tree[1]);
  for (int j = 1, i = k + 1; i <= n; i++, j++) {
    update(1, 1, n, st[j], st[j], -100000000);
    update(1, 1, n, st[i], et[i], 1);
    ans.push_back(tree[1]);
  }
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
}
