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;
int n, k, a[1000005];
struct node {
  int to, nxt, w;
} e[1000005 << 1];
int bian = 0, head[1000005];
inline void add(int a, int b) {
  e[++bian].to = b;
  e[bian].nxt = head[a];
  head[a] = bian;
}
inline void dkr() {
  stack<int> s;
  for (int i = n; i >= 1; i--) {
    while (!s.empty() && a[s.top()] <= a[i]) s.pop();
    if (!s.empty()) {
      int f = s.top();
      add(i, f), add(f, i);
    } else
      add(i, n + 1), add(n + 1, i);
    s.push(i);
  }
}
int in[1000005], out[1000005], t = 0;
inline void dfs(int x, int fa) {
  in[x] = ++t;
  for (int i = head[x]; i; i = e[i].nxt) {
    int T = e[i].to;
    if (T == fa) continue;
    dfs(T, x);
  }
  out[x] = t;
}
struct xds {
  int lazy, l, r, w;
} tre[1000005 << 2];
inline void build(int root, int L, int R) {
  tre[root].l = L, tre[root].r = R;
  if (L == R) return;
  int M = (L + R) >> 1;
  build(root << 1, L, M), build(root << 1 | 1, M + 1, R);
}
inline void pushup(int root) {
  tre[root].w = max(tre[root << 1].w, tre[root << 1 | 1].w);
}
inline void pushdown(int root) {
  if (tre[root].lazy) {
    tre[root << 1].lazy += tre[root].lazy,
        tre[root << 1 | 1].lazy += tre[root].lazy;
    tre[root << 1].w += tre[root].lazy, tre[root << 1 | 1].w += tre[root].lazy;
    tre[root].lazy = 0;
  }
}
inline void update(int root, int L, int R, int pos) {
  if (tre[root].l >= L && tre[root].r <= R) {
    tre[root].w += pos, tre[root].lazy += pos;
    return;
  }
  pushdown(root);
  int mid = (tre[root].l + tre[root].r) >> 1;
  if (L <= mid) update(root << 1, L, R, pos);
  if (R > mid) update(root << 1 | 1, L, R, pos);
  pushup(root);
}
inline int getsum(int root, int L, int R) {
  if (tre[root].l > R || tre[root].r < L) return 0;
  if (tre[root].l >= L && tre[root].r <= R) return tre[root].w;
  pushdown(root);
  int ans = 0, mid = (tre[root].l + tre[root].r) >> 1;
  if (L <= mid) ans = max(ans, getsum(root << 1, L, R));
  if (R > mid) ans = max(ans, getsum(root << 1 | 1, L, R));
  return ans;
}
inline int read() {
  int res = 0, f = 1;
  char c;
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) res = (res << 1) + (res << 3) + (c ^ 48);
  return res * f;
}
inline void write(int x) {
  static int a[35];
  int top = 0;
  if (x < 0) putchar('-'), x = -x;
  do {
    a[top++] = x % 10, x /= 10;
  } while (x);
  while (top) putchar(a[--top] + 48);
  putchar(' ');
}
signed main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  dkr();
  dfs(n + 1, 0);
  build(1, 1, n + 1);
  for (int i = 1; i <= k; i++) update(1, in[i], out[i], 1);
  for (int i = 1; i + k - 1 <= n; i++) {
    int rr = i + k - 1;
    write(getsum(1, 1, n + 1));
    update(1, in[i], out[i], -1);
    update(1, in[rr + 1], out[rr + 1], 1);
  }
  return 0;
}
