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 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
template <class T, class S>
inline void add(T& a, S b) {
  a += b;
  if (a >= mod) a -= mod;
}
template <class T, class S>
inline void sub(T& a, S b) {
  a -= b;
  if (a < 0) a += mod;
}
template <class T, class S>
inline bool chkmax(T& a, S b) {
  return a < b ? a = b, true : false;
}
template <class T, class S>
inline bool chkmin(T& a, S b) {
  return a > b ? a = b, true : false;
}
int n, k, to[N], a[N];
vector<int> G[N];
int in[N], ot[N], idx;
vector<int> ans;
struct SegmentTree {
  int mx[N << 2], lazy[N << 2];
  inline void pull(int rt) { mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]); }
  inline void push(int rt) {
    if (lazy[rt]) {
      mx[rt << 1] += lazy[rt];
      mx[rt << 1 | 1] += lazy[rt];
      lazy[rt << 1] += lazy[rt];
      lazy[rt << 1 | 1] += lazy[rt];
      lazy[rt] = 0;
    }
  }
  void update(int L, int R, int val, int l, int r, int rt) {
    if (R < l || r < L || R < L) return;
    if (L <= l && r <= R) {
      mx[rt] += val;
      lazy[rt] += val;
      return;
    }
    push(rt);
    int mid = l + r >> 1;
    update(L, R, val, l, mid, rt << 1);
    update(L, R, val, mid + 1, r, rt << 1 | 1);
    pull(rt);
  }
  void setVal(int p, int val, int l, int r, int rt) {
    if (l == r) {
      mx[rt] = val;
      return;
    }
    push(rt);
    int mid = l + r >> 1;
    if (p <= mid)
      setVal(p, val, l, mid, rt << 1);
    else
      setVal(p, val, mid + 1, r, rt << 1 | 1);
    pull(rt);
  }
  int query(int L, int R, int l, int r, int rt) {
    if (R < l || r < L || R < L) return 0;
    if (L <= l && r <= R) return mx[rt];
    push(rt);
    int mid = l + r >> 1;
    return max(query(L, R, l, mid, rt << 1),
               query(L, R, mid + 1, r, rt << 1 | 1));
  }
} Tree;
void dfs(int u) {
  in[u] = ++idx;
  for (auto& v : G[u]) dfs(v);
  ot[u] = idx;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  stack<int> stk;
  for (int i = n; i >= 1; i--) {
    while (((int)stk.size()) && a[stk.top()] <= a[i]) stk.pop();
    to[i] = ((int)stk.size()) ? stk.top() : n + 1;
    stk.push(i);
    G[to[i]].push_back(i);
  }
  dfs(n + 1);
  for (int i = n; i > n - k; i--) {
    Tree.setVal(in[i], Tree.query(in[to[i]], in[to[i]], 1, n + 1, 1) + 1, 1,
                n + 1, 1);
  }
  ans.push_back(Tree.mx[1]);
  for (int i = n - k; i >= 1; i--) {
    Tree.update(in[i + k], ot[i + k], -1, 1, n + 1, 1);
    Tree.setVal(in[i], Tree.query(in[to[i]], in[to[i]], 1, n + 1, 1) + 1, 1,
                n + 1, 1);
    ans.push_back(Tree.mx[1]);
  }
  reverse((ans).begin(), (ans).end());
  for (int i = 0; i < ((int)ans.size()); i++)
    printf("%d%c", ans[i], " \n"[i == ((int)ans.size()) - 1]);
  return 0;
}
