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: Time Limit Exceeded)

n, x, y = gets.split.map(&:to_i)
graph = Array.new(n) { Array.new }
# 無向グラフなので両始点のグラフに入れる
0.upto(n - 2) do |i|
  graph[i] << [i, i + 1]
  graph[i + 1] << [i + 1, i]
end
graph[x - 1] << [x - 1, y - 1]
graph[y - 1] << [y - 1, x - 1]
dists = Array.new(n) { Array.new(n, 0) }
0.upto(n - 1) do |i|
  queue = [[i, i, 0]]
  until queue.empty?
    from, to, dist = queue.shift
    graph[to].each do |f, t|
      next if t == from
      if dists[i][t] == 0
        dists[i][t] = dist + 1
        queue << [f, t, dist + 1]
      end
    end
  end
end
dists.flatten!
1.upto(n - 1) do |i|
  puts dists.count(i) / 2
end