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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int inf = 2e9;
const int maxn = 1e6 + 5;
struct MaxLazySegmentTree {
  vector<int> t, o;
  void init(int n) {
    n += 10;
    t.resize(4 * n);
    o.resize(4 * n);
  }
  void push(int v) {
    t[2 * v] += o[v];
    t[2 * v + 1] += o[v];
    o[2 * v] += o[v];
    o[2 * v + 1] += o[v];
    o[v] = 0;
  }
  int queryMax(int v, int tl, int tr, int l, int r) {
    if (l > tr || r < tl) return -inf;
    if (l <= tl && tr <= r) {
      return t[v];
    } else {
      push(v);
      int tm = (tl + tr) / 2;
      return max(queryMax(2 * v, tl, tm, l, r),
                 queryMax(2 * v + 1, tm + 1, tr, l, r));
    }
  }
  void rangeAdd(int v, int tl, int tr, int l, int r, int dx) {
    if (l > r) return;
    if (l > tr || r < tl) return;
    if (l <= tl && tr <= r) {
      t[v] += dx;
      o[v] += dx;
    } else {
      push(v);
      int tm = (tl + tr) / 2;
      rangeAdd(2 * v, tl, tm, l, r, dx);
      rangeAdd(2 * v + 1, tm + 1, tr, l, r, dx);
      t[v] = max(t[2 * v], t[2 * v + 1]);
    }
  }
};
int n, k;
int a[maxn];
int nl[maxn];
int tin[maxn], tout[maxn];
vector<int> g[maxn];
void add_edge(int u, int v) {
  g[u].push_back(v);
  g[v].push_back(u);
}
int cloc = 1;
void dfs(int at, int p) {
  tin[at] = cloc++;
  for (int to : g[at]) {
    if (to == p) continue;
    dfs(to, at);
  }
  tout[at] = cloc++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  a[++n] = inf;
  nl[n] = n;
  for (int i = n - 1; i >= 1; i--) {
    nl[i] = i + 1;
    while (nl[i] < n && a[i] >= a[nl[i]]) {
      nl[i] = nl[nl[i]];
    }
  }
  for (int i = 1; i <= n; i++) {
    g[nl[i]].push_back(i);
  }
  dfs(n, -1);
  int N = cloc;
  MaxLazySegmentTree tree;
  tree.init(N);
  auto upd = [&](int i, int dx) { tree.rangeAdd(1, 1, N, i, N, dx); };
  auto get = [&]() { return tree.queryMax(1, 1, N, 1, N); };
  for (int i = 1; i <= k; i++) {
    upd(tin[i], +1);
    upd(tout[i], -1);
  }
  cout << get() << " ";
  for (int i = k + 1; i < n; i++) {
    upd(tin[i - k], -1);
    upd(tout[i - k], +1);
    upd(tin[i], +1);
    upd(tout[i], -1);
    cout << get() << " ";
  }
  cout << endl;
  return 0;
}
