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>
#pragma GCC optimization("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = 1e6 + 7;
const int INF = 1e9 + 7;
int t[4 * N], lz[4 * N];
int tin[N], tout[N], tim;
int a[N], p[N];
int n, k;
vector<int> g[N];
void dfs(int v) {
  tin[v] = ++tim;
  for (int to : g[v]) dfs(to);
  tout[v] = tim;
}
void push(int v, int l, int r) {
  if (l != r) {
    lz[v + v] += lz[v];
    lz[v + v + 1] += lz[v];
  }
  t[v] += lz[v];
  lz[v] = 0;
}
void upd(int v, int l, int r, int ql, int qr, int val) {
  push(v, l, r);
  if (ql <= l && r <= qr) {
    lz[v] = val;
    push(v, l, r);
    return;
  }
  if (r < ql || qr < l) return;
  int mid = (l + r) >> 1;
  upd(v + v, l, mid, ql, qr, val);
  upd(v + v + 1, mid + 1, r, ql, qr, val);
  t[v] = max(t[v + v], t[v + v + 1]);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  stack<int> st;
  for (int i = n; i >= 1; i--) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (!st.empty()) {
      p[i] = st.top();
      g[p[i]].push_back(i);
    }
    st.push(i);
  }
  for (int i = n; i >= 1; i--)
    if (!p[i]) dfs(i);
  for (int i = 1; i <= k; i++) upd(1, 1, n, tin[i], tout[i], 1);
  printf("%d ", t[1]);
  for (int i = k + 1; i <= n; i++) {
    upd(1, 1, n, tin[i - k], tout[i - k], -1);
    upd(1, 1, n, tin[i], tout[i], 1);
    printf("%d ", t[1]);
  }
}
