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 INF = 200000000000000000LL;
const long double EPS = 1e-9;
const int inf = 0x3f3f3f3f;
struct SegmentTree {
  int Tree[4 * 1000007], Lazy[4 * 1000007];
  SegmentTree() {}
  void pushdown(int at) {
    if (Lazy[at]) {
      Lazy[(at << 1)] += Lazy[at];
      Lazy[((at << 1) | 1)] += Lazy[at];
      Tree[(at << 1)] += Lazy[at];
      Tree[((at << 1) | 1)] += Lazy[at];
      Lazy[at] = 0;
    }
  }
  void build(int at, int l, int r) {
    Lazy[at] = 0;
    if (l == r) {
      Tree[at] = 0;
      return;
    }
    int mid = (l + r) / 2;
    build((at << 1), l, mid);
    build(((at << 1) | 1), mid + 1, r);
    Tree[at] = max(Tree[(at << 1)], Tree[((at << 1) | 1)]);
    Lazy[at] = 0;
  }
  void update(int at, int l, int r, int x, int y, int val) {
    if (x > r || y < l) return;
    if (x <= l && r <= y) {
      Tree[at] += val;
      Lazy[at] += val;
      return;
    }
    if (l != r) pushdown(at);
    int mid = (l + r) / 2;
    update((at << 1), l, mid, x, y, val);
    update(((at << 1) | 1), mid + 1, r, x, y, val);
    Tree[at] = max(Tree[(at << 1)], Tree[((at << 1) | 1)]);
  }
  int query(int at, int l, int r, int x, int y) {
    if (x > r || y < l) return -inf;
    if (x <= l && r <= y) return Tree[at];
    if (l != r) pushdown(at);
    int mid = (l + r) / 2;
    return max(query((at << 1), l, mid, x, y),
               query(((at << 1) | 1), mid + 1, r, x, y));
  }
} seg;
const int N = 1e6 + 7;
vector<int> g[N];
int a[N];
int in[N], out[N], counter = 0;
int n, k;
void dfs(int u) {
  in[u] = ++counter;
  for (int v : g[u]) dfs(v);
  out[u] = counter;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  a[n + 1] = inf;
  stack<int> st;
  st.push(n + 1);
  for (int i = n; i > 0; --i) {
    while (!st.empty() and a[st.top()] <= a[i]) {
      st.pop();
    }
    int pos = st.top();
    g[pos].push_back(i);
    st.push(i);
  }
  counter = 0;
  dfs(n + 1);
  seg.build(1, 1, counter);
  for (int i = 1; i <= n; ++i) {
    seg.update(1, 1, counter, in[i], out[i], 1);
    int ep = i - k;
    if (ep > 0) seg.update(1, 1, counter, in[ep], out[ep], -1);
    if (i >= k) {
      int res = seg.query(1, 1, counter, 1, counter);
      cout << res << " ";
    }
  }
  cout << "\n";
  return 0;
}
