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 = 1000010;
struct SegementTree {
  int mx, lz;
};
SegementTree tr[maxn * 4];
int L[maxn], a[maxn];
void pushup(int x) { tr[x].mx = max(tr[(x << 1)].mx, tr[((x << 1) | 1)].mx); }
void maintain(int x, int val) {
  tr[x].mx += val;
  tr[x].lz += val;
}
void pushdown(int x) {
  if (tr[x].lz) {
    maintain((x << 1), tr[x].lz);
    maintain(((x << 1) | 1), tr[x].lz);
    tr[x].lz = 0;
  }
}
void build(int x, int l, int r) {
  if (l == r) {
    tr[x].mx = 0;
    tr[x].lz = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build((x << 1), l, mid);
  build(((x << 1) | 1), mid + 1, r);
  pushup(x);
}
void update(int x, int l, int r, int ql, int qr, int val) {
  if (l >= ql && r <= qr) {
    maintain(x, val);
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(x);
  if (ql <= mid) update((x << 1), l, mid, ql, qr, val);
  if (qr > mid) update(((x << 1) | 1), mid + 1, r, ql, qr, val);
  pushup(x);
}
int query(int x, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) {
    return tr[x].mx;
  }
  int mid = (l + r) >> 1, ans = 0;
  pushdown(x);
  if (ql <= mid) ans = max(ans, query((x << 1), l, mid, ql, qr));
  if (qr > mid) ans = max(ans, query(((x << 1) | 1), mid + 1, r, ql, qr));
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[0] = 0x3f3f3f3f;
  for (int i = 1; i <= n; i++) {
    L[i] = i - 1;
    while (a[i] > a[L[i]]) {
      L[i] = L[L[i]];
    }
  }
  build(1, 1, n);
  for (int i = 1; i < m; i++) update(1, 1, n, L[i] + 1, i, 1);
  for (int i = 1, j = m; j <= n; i++, j++) {
    update(1, 1, n, L[j] + 1, j, 1);
    printf("%d ", query(1, 1, n, i, j));
  }
}
