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, INF = 2147483647;
struct Edge {
  int to, nex;
} edge[N];
int idx;
int h[N];
void add_edge(int u, int v) {
  edge[++idx] = {v, h[u]};
  h[u] = idx;
}
struct Segment_Tree {
  int l, r;
  int sum;
  int add;
} tree[N << 2];
int dep[N], siz[N];
int id[N], timestamp;
int big[N], stk[N], top;
int a[N], vis[N];
int n, k;
void dfs(int p, int fa) {
  id[p] = ++timestamp;
  siz[p] = 1;
  if (~fa) dep[p] = dep[fa] + 1;
  for (int i = h[p]; ~i; i = edge[i].nex) {
    int to = edge[i].to;
    dfs(to, p);
    siz[p] += siz[to];
  }
}
void push_up(int p) {
  tree[p].sum = max(tree[p << 1].sum, tree[p << 1 | 1].sum);
}
void spread(int p) {
  if (tree[p].add) {
    tree[p << 1].sum += tree[p].add;
    tree[p << 1 | 1].sum += tree[p].add;
    tree[p << 1].add += tree[p].add;
    tree[p << 1 | 1].add += tree[p].add;
    tree[p].add = 0;
  }
}
void build(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  if (l == r) return;
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  push_up(p);
}
void update(int p, int l, int r, int v) {
  if (l <= tree[p].l && r >= tree[p].r) {
    tree[p].sum += v;
    tree[p].add += v;
    return;
  }
  spread(p);
  int mid = tree[p].l + tree[p].r >> 1;
  if (l <= mid) update(p << 1, l, r, v);
  if (r > mid) update(p << 1 | 1, l, r, v);
  push_up(p);
}
int main() {
  scanf("%d%d", &n, &k);
  memset(h, -1, sizeof h);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = n; i; i--) {
    while (top && a[i] >= a[stk[top]]) top--;
    big[i] = stk[top];
    add_edge(stk[top], i);
    stk[++top] = i;
  }
  dfs(0, -1);
  build(1, 1, n + 1);
  for (int i = 1; i <= k - 1; i++) update(1, id[i], id[i] + siz[i] - 1, 1);
  for (int l = 1, r = k; r <= n; l++, r++) {
    update(1, id[r], id[r] + siz[r] - 1, 1);
    printf("%d ", tree[1].sum);
    update(1, id[l], id[l] + siz[l] - 1, -1);
  }
  return 0;
}
