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 (2.7.1) (judge verdict: Accepted)

def bfs(ini, g)
  dist = Array.new(g.length,-1)
  dist[ini] = 0
  queue = [ini]
  while !queue.empty?
    v = queue[0]
    queue.shift
    g[v].each{|des|
      if dist[des] == -1
        dist[des] = dist[v] + 1
        queue.push(des)
      end
     }
   end
   dist
end

n,x,y = gets.chomp.split(" ").map(&:to_i)
g = Array.new(n).map{Array.new}
(n-1).times do |i|
  g[i].push(i+1)
  g[i+1].push(i)
end
g[x-1].push(y-1)
g[y-1].push(x-1)
ans = Array.new(n,0)
n.times do |i|
  bfs(i,g).each do |z|
    ans[z] += 1
  end
end

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