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 = 1000005;
const int MAXSEG = 1 << 21;
const int INFTY = 1 << 29;
int N, K;
int A[MAXN];
vector<int> adj[MAXN];
int deg[MAXN];
int R[MAXN][2], ind = 0;
int seg[MAXSEG];
int lazy[MAXSEG];
void dfs(int cur) {
  R[cur][0] = ++ind;
  for (auto nxt : adj[cur]) dfs(nxt);
  R[cur][1] = ind;
}
void down(int node) {
  if (lazy[node] != 0) {
    seg[2 * node] += lazy[node];
    seg[2 * node + 1] += lazy[node];
    lazy[2 * node] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
    lazy[node] = 0;
  }
}
void range_update(int node, int left, int right, int ql, int qr) {
  if (left > right || ql > qr || right < ql || qr < left) return;
  if (ql <= left && right <= qr) {
    seg[node]++;
    lazy[node]++;
    return;
  }
  down(node);
  int mid = (left + right) / 2;
  range_update(2 * node, left, mid, ql, qr);
  range_update(2 * node + 1, mid + 1, right, ql, qr);
  seg[node] = max(seg[2 * node], seg[2 * node + 1]);
}
void point_update(int node, int left, int right, int idx) {
  if (left == idx && idx == right) {
    seg[node] = -INFTY;
    return;
  }
  down(node);
  int mid = (left + right) / 2;
  if (idx <= mid)
    point_update(2 * node, left, mid, idx);
  else
    point_update(2 * node + 1, mid + 1, right, idx);
  seg[node] = max(seg[2 * node], seg[2 * node + 1]);
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  vector<int> stk;
  for (int i = N; i >= 1; i--) {
    while (!stk.empty() && A[i] >= A[stk.back()]) stk.pop_back();
    if (!stk.empty()) {
      adj[stk.back()].push_back(i);
      deg[i]++;
    }
    stk.push_back(i);
  }
  for (int i = 1; i <= N; i++)
    if (deg[i] == 0) dfs(i);
  for (int i = 1; i <= K; i++) range_update(1, 1, N, R[i][0], R[i][1]);
  printf("%d", seg[1]);
  for (int i = 2; i + K - 1 <= N; i++) {
    point_update(1, 1, N, R[i - 1][0]);
    range_update(1, 1, N, R[i + K - 1][0], R[i + K - 1][1]);
    printf(" %d", seg[1]);
  }
  printf("\n");
  return 0;
}
