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)

X -= 1
Y -= 1

ans = Array.new(N, 0)

N.times do |sv|
  depths = Array.new(N, Float::INFINITY)
  queue = Queue.new

  depths[sv] = 0
  queue.push sv

  until queue.empty? do
    v = queue.pop
    d = depths[v]

    if v + 1 < N && depths[v + 1] == Float::INFINITY
      depths[v+1] = d + 1
      queue.push v + 1
    end

    if v - 1 >= 0 && depths[v - 1] == Float::INFINITY
      depths[v-1] = d + 1
      queue.push v - 1
    end

    if v == X
      if depths[Y] == Float::INFINITY
        depths[Y] = d + 1
        queue.push Y
      end
    end

    if v == Y
      if depths[X] == Float::INFINITY
        depths[X] = d + 1
        queue.push X
      end
    end
  end

  N.times do |i|
    ans[depths[i]] += 1
  end
end

ans[1..(N-1)].each do |a|
  puts a / 2
end