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 MAX_N = 1000025;
int in[MAX_N], out[MAX_N], arr[MAX_N], dep, s[MAX_N << 2], col[MAX_N << 2],
    ans[MAX_N];
stack<int> st;
vector<int> vt[MAX_N];
void dfs(int rt, int fa) {
  in[rt] = ++dep;
  for (int i = 0; i < vt[rt].size(); i++) {
    int to = vt[rt][i];
    if (to == fa) continue;
    dfs(to, rt);
  }
  out[rt] = dep;
}
void up(int rt) { s[rt] = max(s[rt << 1], s[rt << 1 | 1]); }
void build(int rt, int l, int r) {
  col[rt] = 0;
  if (l == r) {
    s[rt] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  up(rt);
}
void down(int rt, int l, int r) {
  if (col[rt]) {
    int mid = (l + r) >> 1;
    col[rt << 1] += col[rt];
    col[rt << 1 | 1] += col[rt];
    s[rt << 1] += col[rt];
    s[rt << 1 | 1] += col[rt];
    col[rt] = 0;
  }
}
void update(int rt, int l, int r, int x, int y, int v) {
  if (x <= l && r <= y) {
    col[rt] += v;
    s[rt] += v;
    return;
  }
  int mid = (l + r) >> 1;
  down(rt, l, r);
  if (x <= mid) update(rt << 1, l, mid, x, y, v);
  if (mid < y) update(rt << 1 | 1, mid + 1, r, x, y, v);
  up(rt);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
  build(1, 1, n + 1);
  for (int i = n; i >= 1; i--) {
    while (!st.empty() && arr[st.top()] <= arr[i]) st.pop();
    if (st.empty()) {
      vt[0].push_back(i);
    } else {
      vt[st.top()].push_back(i);
    }
    st.push(i);
  }
  dfs(0, -1);
  for (int i = n - m + 1; i <= n; ++i) update(1, 1, n + 1, in[i], out[i], 1);
  ans[n - m + 1] = s[1];
  for (int i = n - m; i >= 1; i--) {
    update(1, 1, n + 1, in[i + m], out[i + m], -1);
    update(1, 1, n + 1, in[i], out[i], 1);
    ans[i] = s[1];
  }
  for (int i = 1; i <= n - m + 1; ++i)
    i == n - m + 1 ? printf("%d\n", ans[i]) : printf("%d ", ans[i]);
  return 0;
}
