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 MAXN = 1e6 + 5;
const int MAXLog = 20;
struct SegmentTree {
  int maxVal, lazy;
  int l, r;
  SegmentTree *L, *R;
  SegmentTree() {}
  SegmentTree(int l, int r) {
    this->maxVal = this->lazy = 0;
    this->l = l;
    this->r = r;
    this->L = this->R = nullptr;
  }
  void build() {
    if (l == r) return;
    L = new SegmentTree(l, (l + r) / 2);
    R = new SegmentTree((l + r) / 2 + 1, r);
    L->build();
    R->build();
  }
  void reset() {
    maxVal = lazy = 0;
    if (L != nullptr) L->reset();
    if (R != nullptr) R->reset();
  }
  void updateLazy() {
    maxVal += lazy;
    if (L != nullptr) {
      L->lazy += lazy;
      R->lazy += lazy;
    }
    lazy = 0;
  }
  int query(int ql, int qr) {
    updateLazy();
    if (l >= ql && qr <= r) return maxVal;
    if (r < ql || l > qr) return 0;
    return max(L->query(ql, qr), R->query(ql, qr));
  }
  void update(int ql, int qr, int val) {
    updateLazy();
    if (l >= ql && r <= qr) {
      lazy += val;
      updateLazy();
      return;
    }
    if (r < ql || l > qr) return;
    L->update(ql, qr, val);
    R->update(ql, qr, val);
    maxVal = max(L->maxVal, R->maxVal);
  }
};
int n, k;
int a[MAXN];
int nxt[MAXN];
SegmentTree *T;
int orderInd[MAXN], treeSize[MAXN];
vector<int> graph[MAXN], dfsOrder;
void DFS(int x) {
  orderInd[x] = dfsOrder.size();
  dfsOrder.push_back(x);
  treeSize[x] = 1;
  for (int y : graph[x]) {
    DFS(y);
    treeSize[x] += treeSize[y];
  }
}
void init() {
  stack<int> st;
  for (int i = n - 1; i >= 0; i--) {
    while (st.empty() == false && a[st.top()] <= a[i]) st.pop();
    nxt[i] = ((st.empty() == true) ? n : st.top());
    st.push(i);
    graph[nxt[i]].push_back(i);
  }
  DFS(n);
  T = new SegmentTree(0, dfsOrder.size() - 1);
  T->build();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  vector<int> output;
  double average = 0;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  init();
  for (int i = n - 1; i >= n - k; i--)
    T->update(orderInd[i], orderInd[i] + treeSize[i] - 1, +1);
  output.push_back(T->query(0, dfsOrder.size() - 1));
  for (int i = n - 2; i >= k - 1; i--) {
    T->update(orderInd[i - k + 1],
              orderInd[i - k + 1] + treeSize[i - k + 1] - 1, +1);
    T->update(orderInd[i + 1], orderInd[i + 1] + treeSize[i + 1] - 1, -1);
    output.push_back(T->query(0, dfsOrder.size() - 1));
  }
  reverse(output.begin(), output.end());
  for (int x : output) cout << x << " ";
  cout << '\n';
}
