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.chomp.split.map(&:to_i)

def calc(dp, x, y, d)
  return if x < 0 || y >= N

  return if dp[x][y] <= d

  dp[x][y] = d

  calc(dp, x - 1, y, d + 1)

  calc(dp, x + 1, y, d + 1)

  calc(dp, x, y + 1, d + 1)

  calc(dp, x, y - 1, d + 1)
end

dp = Array.new(N) { Array.new(N, 0) }

0.upto(N - 2) do |i|
  (i + 1).upto(N - 1) do |j|
    dp[i][j] = j - i
  end
end

calc(dp, X - 1, Y - 1, 1)

count = {}

0.upto(N - 2) do |i|
  (i + 1).upto(N - 1) do |j|
    count[dp[i][j]] ||= 0
    count[dp[i][j]] += 1
  end
end

1.upto(N - 1) do |k|
  puts count[k] || 0
end