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 = 1000005;
int n, m, a[N], dp[N], fi[N], l, ne[N], zz[N], rk[N], size[N], tot;
struct Tree {
  int l, r, num, flag;
} T1[N * 4], T2[N * 4];
void jb(int x, int y) {
  ne[++tot] = fi[x];
  fi[x] = tot;
  zz[tot] = y;
}
void build1(int x, int l, int r) {
  T1[x].l = l;
  T1[x].r = r;
  T1[x].num = 1e9;
  if (l == r) return;
  int mid = (l + r) / 2;
  build1(x * 2, l, mid);
  build1(x * 2 + 1, mid + 1, r);
}
void build2(int x, int l, int r) {
  T2[x].l = l;
  T2[x].r = r;
  if (l == r) return;
  int mid = (l + r) / 2;
  build2(x * 2, l, mid);
  build2(x * 2 + 1, mid + 1, r);
}
void insert1(int x, int y, int z) {
  T1[x].num = z;
  if (T1[x].l == T1[x].r) return;
  int mid = (T1[x].l + T1[x].r) / 2;
  if (y <= mid)
    insert1(x * 2, y, z);
  else
    insert1(x * 2 + 1, y, z);
}
int find1(int x, int l, int r) {
  if (T1[x].l > r || l > T1[x].r) return 1e9;
  if (T1[x].l >= l && T1[x].r <= r) return T1[x].num;
  return min(find1(x * 2, l, r), find1(x * 2 + 1, l, r));
}
void down(int x) {
  T2[x * 2].num += T2[x].flag;
  T2[x * 2 + 1].num += T2[x].flag;
  T2[x * 2].flag += T2[x].flag;
  T2[x * 2 + 1].flag += T2[x].flag;
  T2[x].flag = 0;
}
void insert2(int x, int l, int r) {
  if (T2[x].l > r || l > T2[x].r) return;
  if (T2[x].l >= l && T2[x].r <= r) {
    T2[x].num++;
    T2[x].flag++;
    return;
  }
  down(x);
  insert2(x * 2, l, r);
  insert2(x * 2 + 1, l, r);
  T2[x].num = max(T2[x * 2].num, T2[x * 2 + 1].num);
}
void del(int x, int y) {
  if (T2[x].l == T2[x].r) {
    T2[x].num = -1e9;
    return;
  }
  down(x);
  int mid = (T1[x].l + T1[x].r) / 2;
  if (y <= mid)
    del(x * 2, y);
  else
    del(x * 2 + 1, y);
  T2[x].num = max(T2[x * 2].num, T2[x * 2 + 1].num);
}
void dfs(int x, int y) {
  size[x] = 1;
  rk[x] = ++l;
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != y) dfs(zz[i], x), size[x] += size[zz[i]];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build1(1, 1, n + 1);
  insert1(1, n + 1, n + 1);
  for (int i = n; i; i--) {
    dp[i] = find1(1, a[i] + 1, n + 1);
    insert1(1, a[i], i);
    jb(dp[i], i);
  }
  dfs(n + 1, 0);
  build2(1, 1, n + 1);
  for (int i = 1; i <= m; i++) insert2(1, rk[i], size[i] + rk[i] - 1);
  printf("%d ", T2[1].num);
  for (int i = m + 1; i <= n; i++) {
    del(1, rk[i - m]);
    insert2(1, rk[i], size[i] + rk[i] - 1);
    printf("%d ", T2[1].num);
  }
  return 0;
}
