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 comment(linker, "/STACK:16000000")
using namespace std;
const int Maxn = 1000005;
const int Maxm = 4194304;
int n, k;
vector<int> neigh[Maxn];
int cur, lef[Maxn], rig[Maxn];
int st[Maxm], flag[Maxm];
void Traverse(int v) {
  lef[v] = ++cur;
  for (int i = 0; i < neigh[v].size(); i++) Traverse(neigh[v][i]);
  rig[v] = cur;
}
void downOn(int v, int f) {
  st[v] += f;
  flag[v] += f;
}
void Down(int v) {
  if (flag[v]) {
    downOn(2 * v, flag[v]);
    downOn(2 * v + 1, flag[v]);
    flag[v] = 0;
  }
}
void Union(int v) { st[v] = max(st[2 * v], st[2 * v + 1]); }
void Update(int v, int l, int r, int a, int b, int val) {
  if (l == a && r == b)
    downOn(v, val);
  else {
    int m = l + r >> 1;
    Down(v);
    if (a <= m) Update(2 * v, l, m, a, min(m, b), val);
    if (m + 1 <= b) Update(2 * v + 1, m + 1, r, max(m + 1, a), b, val);
    Union(v);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  vector<pair<int, int> > S;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    while (!S.empty() && S.back().first < a) {
      neigh[i].push_back(S.back().second);
      S.pop_back();
    }
    S.push_back(pair<int, int>(a, i));
  }
  for (int i = n; i > 0; i--)
    if (!lef[i]) Traverse(i);
  for (int i = 1; i < k; i++) Update(1, 1, n, lef[i], rig[i], 1);
  for (int i = k; i <= n; i++) {
    Update(1, 1, n, lef[i], rig[i], 1);
    printf("%d%c", st[1], i + 1 <= n ? ' ' : '\n');
    Update(1, 1, n, lef[i - (k - 1)], rig[i - (k - 1)], -1);
  }
  return 0;
}
