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: Wrong Answer)

n,x,y,c = gets.split.map {|i| i.to_i-1 }
x,y = n-y,n-x  if x > n-y
a = Array.new n+1,0
f =->(i,j,k) {
  begin
    s = 0
    s += 1  if j > i+=1
    s += 1  if i < j-=1
    s += 1  if n > k+=1
    a[c+=1] += s
  end while s > 0
}
y.times {|i|
  c = 0
  if x < i
    [y-i,i-x].min.times { a[c+=1] += 1; i+=1 }
    if i < y-1
      a[c+=1] += 2
      f[i+1,y,y-1]
    else
      (n-i).times { a[c+=1] += 1 }
    end
  else
    (x-i).times { a[c+=1] += 1 }
    a[c+=1] += 2
    f[x+1,y,y-1]
  end
}
(m=n-y).times {|i| a[m-i] += i+1 }
p a[1,n]