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 + 10;
int n, k, a[N];
vector<int> g[N];
int sta[N], top, dep[N];
int dfn[N], dfn_end[N], clk;
int val[N];
void dfs(int u) {
  dfn[u] = ++clk;
  val[clk] = dep[u];
  for (int v : g[u]) {
    dfs(v);
  }
  dfn_end[u] = clk;
}
int mx[N * 10], tag[N * 10];
const int inf = -0xcfcfcfcf;
void push(int id) {
  if (tag[id]) {
    tag[(id << 1)] += tag[id], tag[(id << 1 | 1)] += tag[id];
    mx[id] += tag[id], tag[id] = 0;
  }
}
void upd(int id) {
  push((id << 1)), push((id << 1 | 1));
  mx[id] = max(mx[(id << 1)], mx[(id << 1 | 1)]);
}
void modify(int id, int l, int r, int ql, int qr, int val) {
  int mid = (l + r) >> 1;
  push(id);
  if (ql <= l && r <= qr) {
    tag[id] += val;
    return;
  } else if (qr <= mid) {
    modify((id << 1), l, mid, ql, qr, val);
  } else if (ql >= mid + 1) {
    modify((id << 1 | 1), mid + 1, r, ql, qr, val);
  } else {
    modify((id << 1), l, mid, ql, mid, val);
    modify((id << 1 | 1), mid + 1, r, mid + 1, qr, val);
  }
  upd(id);
}
int ans[N];
int main() {
  memset(mx, 0xcf, sizeof mx);
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = n; i; --i) {
    while (top && a[sta[top]] <= a[i]) --top;
    dep[i] = dep[sta[top]] + 1;
    g[sta[top]].push_back(i);
    sta[++top] = i;
  }
  dfs(0);
  int L = n - k + 1, R = n;
  for (int i = L; i <= R; ++i) {
    modify(1, 1, clk, dfn[i], dfn[i], inf + dep[i]);
  }
  for (int i = 1; i <= n - k + 1; ++i) {
    push(1), ans[i] = mx[1];
    modify(1, 1, clk, dfn[R], dfn_end[R], -1);
    modify(1, 1, clk, dfn[L - 1], dfn[L - 1], inf + dep[L - 1]);
    --R, --L;
  }
  for (int i = n - k + 1; i; --i) {
    printf("%d ", ans[i]);
  }
}
