Problem:
Score : 400 points

Problem Statement

We have an undirected graph G with N vertices numbered 1 to N and N edges as follows:

For each i=1,2,...,N-1, there is an edge between Vertex i and Vertex i+1.

There is an edge between Vertex X and Vertex Y.

For each k=1,2,...,N-1, solve the problem below:

Find the number of pairs of integers (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j in G is k.

Constraints

3 \leq N \leq 2 \times 10^3

1 \leq X,Y \leq N

X+1 < Y

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X Y

Output

For each k=1, 2, ..., N-1 in this order, print a line containing the answer to the problem.

Sample Input 1

5 2 4

Sample Output 1

5
4
1
0

The graph in this input is as follows:

There are five pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 1: (1,2)\,,(2,3)\,,(2,4)\,,(3,4)\,,(4,5).

There are four pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 2: (1,3)\,,(1,4)\,,(2,5)\,,(3,5).

There is one pair (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 3: (1,5).

There are no pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 4.

Sample Input 2

3 1 3

Sample Output 2

3
0

The graph in this input is as follows:

Sample Input 3

7 3 7

Sample Output 3

7
8
4
2
0
0

Sample Input 4

10 4 8

Sample Output 4

10
12
10
8
4
1
0
0
0

Solution in Ruby (2.3.3) (judge verdict: Accepted)

n, x, y = gets.chomp.split.map(&:to_i)


# 各点からの最短距離を求める
graph = Hash.new{ |h,k| h[k] = []}
# graph = Array.new(n+1){Array.new}

1.upto(n) do |i|
  graph[i] << i+1 unless i == n
  graph[i] << i-1 unless i == 1
end

graph[x] << y
graph[y] << x



# ans = Hash.new(0)
# arrayなら0を気にしなくて良い
ans = Array.new(n,0)

1.upto(n) do |i|
  # visited = Hash.new(false)

  # queueに距離を持たず、それ用の配列を用意する
  # これにvisitedの役割も持たせている
  dist = Array.new(n+1, -1)

  dist[i] = 0
  queue = [i]

  while queue.size > 0
    v = queue.shift

    graph[v].each do |nv|
      next unless dist[nv] == -1
      dist[nv] = dist[v] + 1
      queue << nv
      ans[dist[nv]] += 1
    end
  end
end


1.upto(n-1) do |i|
  puts ans[i] / 2
end