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 MXN = 1e6 + 5;
const int OFF = 1 << 20;
int timer;
int arr[MXN];
int sol[MXN];
int tin[MXN];
int tout[MXN];
int tree[2 * OFF];
int prop[2 * OFF];
vector<int> adj[MXN];
void update(int l, int r, int dx, int lo = 0, int hi = OFF, int x = 0) {
  if (l <= lo && hi <= r) {
    tree[x] += dx;
    prop[x] += dx;
    return;
  }
  int mid = (lo + hi) / 2;
  if (l < mid) update(l, r, dx, lo, mid, 2 * x + 1);
  if (mid < r) update(l, r, dx, mid, hi, 2 * x + 2);
  tree[x] = max(tree[2 * x + 1], tree[2 * x + 2]) + prop[x];
}
void dfs(int x) {
  tin[x] = timer++;
  for (int y : adj[x]) {
    dfs(y);
  }
  tout[x] = timer;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  stack<int> stk;
  for (int i = n - 1; i >= 0; --i) {
    while (stk.size() && arr[stk.top()] <= arr[i]) {
      stk.pop();
    }
    if (stk.size()) {
      adj[stk.top()].push_back(i);
    }
    stk.push(i);
  }
  for (int i = n - 1; i >= 0; --i) {
    if (!tin[i]) {
      dfs(i);
    }
  }
  update(0, OFF, -MXN);
  for (int i = n - 1; i >= 0; --i) {
    if (n - i > k) {
      int j = i + k;
      update(tin[j], tout[j], -1);
      update(tin[j], tin[j] + 1, -MXN);
    }
    update(tin[i], tout[i], 1);
    update(tin[i], tin[i] + 1, MXN);
    sol[i] = tree[0];
  }
  for (int i = 0; i <= n - k; ++i) {
    cout << sol[i] << ' ';
  }
}
