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 = 1000000 + 10;
int n, k, a[N];
int mx[N << 2], lz[N << 2], pre[N];
int stk[N], top;
void pushdown(int rt) {
  if (lz[rt]) {
    mx[rt << 1] = mx[rt << 1] + lz[rt];
    lz[rt << 1] = lz[rt << 1] + lz[rt];
    mx[rt << 1 | 1] = mx[rt << 1 | 1] + lz[rt];
    lz[rt << 1 | 1] = lz[rt << 1 | 1] + lz[rt];
    lz[rt] = 0;
  }
}
void update(int l, int r, int rt, int L, int R) {
  if (L <= l && r <= R) {
    lz[rt]++;
    mx[rt]++;
    return;
  }
  pushdown(rt);
  int mid = (l + r) >> 1;
  if (L <= mid) update(l, mid, rt << 1, L, R);
  if (R > mid) update(mid + 1, r, rt << 1 | 1, L, R);
  mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}
int query(int l, int r, int rt, int L, int R) {
  if (L <= l && r <= R) {
    return mx[rt];
  }
  pushdown(rt);
  int mid = (l + r) >> 1;
  int ans = 0;
  if (L <= mid) ans = max(ans, query(l, mid, rt << 1, L, R));
  if (R > mid) ans = max(ans, query(mid + 1, r, rt << 1 | 1, L, R));
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    while (top && a[stk[top]] < a[i]) --top;
    pre[i] = stk[top];
    stk[++top] = i;
  }
  for (int i = 1; i <= n; i++) {
    int L = max(pre[i] + 1, i - k + 1);
    int R = i;
    update(1, n, 1, L, R);
    if (i >= k) {
      int lef = max(i - k + 1, 1);
      printf("%d ", query(1, n, 1, lef, R));
    }
  }
}
