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 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize ("O3")
#pragma GCC optimize("unroll-loops")
#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll

int32_t main()
{
	#ifdef ONLINE_JUDGE
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	#endif
	
	int n,x,y;
	cin>>n>>x>>y;
	int dist[n+1][n+1];
	dist[x][y]=1;
	for(int i=1;i<=n;i++)
	{
		//if(i==x or i==y)continue;
		dist[i][x]=dist[x][i]=abs(x-i);
		dist[i][y]=dist[y][i]=abs(y-i);
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=i+1;j<=n;j++)
		{
			dist[i][j]=dist[j][i]=min(abs(i-j),dist[i][x]+dist[y][j]+1);
		}
	}
	vector<int> f(n);
	for(int i=1;i<=n;i++)
	for(int j=i+1;j<=n;j++)
	{f[dist[i][j]]++;}
	for(int i=1;i<n;i++)cout<<f[i]<<'\n';
	
}