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 MOD = 1000000007;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
const int MAXN = 1000005;
int N, K, A[MAXN], par[MAXN], in[MAXN], out[MAXN], cnt;
long long tree[4 * MAXN], lazy[4 * MAXN];
vector<int> v[MAXN];
inline void prop(int rt) {
  tree[rt << 1] += lazy[rt], tree[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 ql, int qr, int rt, long long v) {
  if (l == ql && r == qr) {
    tree[rt] += v, lazy[rt] += v;
    return;
  }
  if (lazy[rt]) prop(rt);
  int mid = (l + r) / 2;
  if (qr <= mid)
    update(l, mid, ql, qr, rt << 1, v);
  else if (ql > mid)
    update(mid + 1, r, ql, qr, rt << 1 | 1, v);
  else
    update(l, mid, ql, mid, rt << 1, v),
        update(mid + 1, r, mid + 1, qr, rt << 1 | 1, v);
  tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);
}
void dfs(int u) {
  in[u] = ++cnt;
  for (int ck : v[u]) dfs(ck);
  out[u] = cnt;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> K;
  for (int i = 1; i <= N; i++) cin >> A[i];
  stack<pair<int, int> > second;
  for (int i = N; i >= 1; i--) {
    while (second.size() && second.top().first <= A[i]) second.pop();
    if (second.empty())
      par[i] = i;
    else
      par[i] = second.top().second, v[second.top().second].push_back(i);
    second.push(make_pair(A[i], i));
  }
  for (int i = 1; i <= N; i++)
    if (par[i] == i) dfs(i);
  for (int i = 1; i <= N; i++) update(1, cnt, in[i], in[i], 1, -INT_MAX);
  for (int i = 1; i <= N; i++) {
    update(1, cnt, in[i], in[i], 1, INT_MAX);
    update(1, cnt, in[i], out[i], 1, 1);
    if (i >= K) {
      cout << tree[1] << " ";
      update(1, cnt, in[i - K + 1], in[i - K + 1], 1, -INT_MAX);
    }
  }
}
