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 + 100;
const int M = 1e6;
const long long mod = 998244353;
const long long MOD = 998244353;
const long double eps = 0.000000001;
const int P = 1336;
const int inf = 1e9 + 7;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
vector<int> no, t, add;
int tin[N], tout[N], tim = -1;
int a[N];
vector<int> g[N];
int n, k;
void push(int v) {
  t[v * 2] += add[v];
  t[v * 2 + 1] += add[v];
  add[v * 2] += add[v];
  add[v * 2 + 1] += add[v];
  add[v] = 0;
}
void upd(int l, int r, int v, int cl, int cr, int zn) {
  if (cr < l || r < cl) return;
  if (l <= cl && cr <= r) {
    t[v] += zn;
    add[v] += zn;
    return;
  }
  if (cl != cr) push(v);
  int mid = (cl + cr) / 2;
  upd(l, r, v * 2, cl, mid, zn);
  upd(l, r, v * 2 + 1, mid + 1, cr, zn);
  t[v] = max(t[v * 2], t[v * 2 + 1]);
}
int maxx(int l, int r, int v, int cl, int cr) {
  if (cr < l || r < cl) return -inf;
  if (l <= cl && cr <= r) return t[v];
  if (cl != cr) push(v);
  int mid = (cl + cr) / 2;
  return max(maxx(l, r, v * 2, cl, mid), maxx(l, r, v * 2 + 1, mid + 1, cr));
}
void update(int v, int zn) {
  no[v] = zn;
  v /= 2;
  while (v) {
    no[v] = min(no[v * 2], no[v * 2 + 1]);
    v /= 2;
  }
}
int minn(int l, int r, int v, int cl, int cr) {
  if (cr < l || r < cl) return n;
  if (l <= cl && cr <= r) return no[v];
  int mid = (cl + cr) / 2;
  return min(minn(l, r, v * 2, cl, mid), minn(l, r, v * 2 + 1, mid + 1, cr));
}
void DFS(int v) {
  tim++;
  tin[v] = tim;
  for (auto to : g[v]) {
    DFS(to);
  }
  tim++;
  tout[v] = tim;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long pr = 1;
  while (pr < n) pr *= 2;
  no.resize(2 * pr);
  for (int i = 0; i < pr + pr; i++) {
    no[i] = n;
  }
  for (int i = n - 1; i >= 0; i--) {
    int x = minn(a[i], n - 1, 1, 0, pr - 1);
    g[x].push_back(i);
    update(pr + a[i] - 1, i);
  }
  DFS(n);
  pr *= 2;
  t.resize(2 * pr);
  add.resize(2 * pr);
  for (int i = 0; i < k; i++) {
    upd(tin[i], tout[i] - 1, 1, 0, pr - 1, 1);
  }
  cout << t[1] << " ";
  for (int i = k; i < n; i++) {
    upd(tin[i - k], tout[i - k] - 1, 1, 0, pr - 1, -1);
    upd(tin[i], tout[i] - 1, 1, 0, pr - 1, 1);
    cout << t[1] << " ";
  }
}
