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)

 #include <stdio.h>
 #include <math.h>


int i, j, k;
int g[2*1000][2*1000];
int d[2000];
int min(int a, int b){
  if(a>b)return b;
  else return a;
}

int abs(int a){
  if (a<0)a*=-1;
  return a;
}

int main(){
   int n,a,b;
   scanf("%d %d %d", &n, &a, &b);
   for(i=0;i<n-1;i++)g[i][i+1]=g[i+1][i]=1;
   for(i=1;i<n+1;i++)d[i]=0;
   g[a-1][b-1]=g[b-1][a-1]=1;


   for(i=1;i<n;i++){
     for(j=i+1;j<n+1;j++){
       d[min(j-i, abs(a-i)+abs(b-j)+1)]+=1;
     }
   }

   for(i=1;i<n;i++)printf("%d\n", d[i]);


   return 0;
 }