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: Accepted)

N, X, Y = gets.split(" ").map(&:to_i)
ans = Array.new(N - 1, 0)

(N - 1).times do |i|
    j = i + 1
    while j < N do
        if X < i + 1 && j + 1 < Y
            min = (((2 * i - X + Y - 2 * j + 1) >= 0) ? (j - i - 1) : (- X + Y + i - j))
            ans[min] += 1
        elsif X < i + 1 && j + 1 >= Y
            min = (((2 * i - X - Y + 3) >= 0) ? (j - i - 1) : (i + j - X - Y + 2))
            ans[min] += 1
        elsif X >= i + 1 && Y >= j + 1
            min = (((X + Y - 2 * j - 1) >= 0) ? (j - i - 1) : (X + Y - i - j - 2))
            ans[min] += 1
        elsif X >= i + 1 && Y < j + 1
            ans[X - Y - i + j] += 1
        end
        j += 1
    end
end

puts ans