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;
int rd() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f * x;
}
struct node {
  int mx, tag;
} tree[1000005 << 2];
int n, k, a[1000005], stk[1000005], tp;
vector<int> G[1000005];
int dfn[1000005], tim, siz[1000005];
void dfs(int u) {
  dfn[u] = ++tim;
  siz[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    dfs(v);
    siz[u] += siz[v];
  }
}
void PushDown(int i) {
  if (tree[i].tag) {
    tree[i << 1].mx += tree[i].tag;
    tree[i << 1].tag += tree[i].tag;
    tree[i << 1 | 1].mx += tree[i].tag;
    tree[i << 1 | 1].tag += tree[i].tag;
    tree[i].tag = 0;
  }
}
void PushUp(int i) { tree[i].mx = max(tree[i << 1].mx, tree[i << 1 | 1].mx); }
void Update(int i, int l, int r, int ql, int qr, int val) {
  if (ql <= l && r <= qr) {
    tree[i].mx += val;
    tree[i].tag += val;
    return;
  }
  PushDown(i);
  int mid = (l + r) >> 1;
  if (ql <= mid) Update(i << 1, l, mid, ql, qr, val);
  if (qr > mid) Update(i << 1 | 1, mid + 1, r, ql, qr, val);
  PushUp(i);
}
int main() {
  n = rd(), k = rd();
  for (int i = 1; i <= n; i++) a[i] = rd();
  for (int i = 1; i <= n; i++) {
    while (tp && a[stk[tp]] < a[i]) {
      G[i].push_back(stk[tp]);
      tp--;
    }
    stk[++tp] = i;
  }
  while (tp) {
    G[n + 1].push_back(stk[tp]);
    tp--;
  }
  dfs(n + 1);
  for (int i = 1; i <= k; i++)
    Update(1, 1, n + 1, dfn[i], dfn[i] + siz[i] - 1, 1);
  printf("%d ", tree[1].mx);
  for (int i = k + 1; i <= n; i++) {
    Update(1, 1, n + 1, dfn[i], dfn[i] + siz[i] - 1, 1);
    Update(1, 1, n + 1, dfn[i - k], dfn[i - k] + siz[i - k] - 1, -1);
    printf("%d ", tree[1].mx);
  }
  return 0;
}
