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: Wrong Answer)

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

h = {}
(n-1).times do |i|
  h[i+1] ||= {}
  h[i+1][i+2] = 1
end

h[x][y] = 1

q = []
q.push([1, 2])

visited = {}

while !q.empty?
  prev, current = q.shift
  visited[prev] ||= {}
  visited[prev][current] = true
  h[prev][current] = [h[prev][current], 1].min

  (1..current-1).each do |i|
    if current == 1
    else
      if h[i][prev].nil?
      else
        if h[i][current]
          h[i][current] = [h[i][prev] + 1, h[i][current]].min
        else
          h[i][current] = h[i][prev] + 1
        end
      end
    end
  end

  if current == n
    break
  end
  h[current].keys.each do |i|
    visited[current] ||= {}
    unless visited[current][i]
      q.push([current, i])
    end
  end
end

ans = {}
(n-1).times do |i|
  i += 1
  h[i].each do |key, val|
    ans[val] ||= 0
    ans[val] += 1
  end
end

(n-1).times do |i|
  puts ans[i+1] || 0
end