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 C (GCC 5.4.1) (judge verdict: Accepted)

main()
{
	int N,X,Y;
	scanf("%d%d%d",&N,&X,&Y);
	int loop=Y-X+1;
	for(int k=1;k<N;k++)
	{
		int ans=0;
		int z=X<k?X-1:k-1;
		if(k-z<=loop/2)
		{
			int t=2*k-loop-1>0?2*k-loop-1:0;
			ans+=(z-t/2)*2;
			ans-=!(loop&1)&&loop/2<k&&k-loop/2<X;
		}
		z=N-Y<k-1?N-Y:k-1;
		if(k-z<=loop/2)
		{
			int t=2*k-loop-1>0?2*k-loop-1:0;
			ans+=(z-t/2)*2;
			ans-=!(loop&1)&&loop/2<k&&k-loop/2<N+1-Y;
		}
		if(N+1+X-Y>k)ans+=N+1+X-Y-k;
		if(k==1)ans--;
		else
		{
			ans-=X>=k;
			ans-=N-Y>=k-1;
		}
		if(k*2<loop)ans+=loop;
		else if(k*2==loop)ans+=loop/2;
		printf("%d\n",ans);
	}
}