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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
#define int long long
#define double long double
#define fo(a,b) for(int a=0;a<b;a++)
#define Sort(a) sort(a.begin(),a.end())
#define rev(a) reverse(a.begin(),a.end())
#define fi first
#define se second
#define sz size()
#define bgn begin()
#define en end()
#define pb push_back
#define pp() pop_back()
#define V vector
#define P pair
#define yuko(a) setprecision(a)
#define uni(a) a.erase(unique(a.begin(),a.end()),a.end())
#define Q queue
#define pri priority_queue
#define Pri priority_queue<int,vector<int>,greater<int>>
#define PriP priority_queue<P<int,int>,vector<P<int,int>>,greater<P<int,int>>>
#define ff first.first
#define fs first.second
#define sf second.first
#define ss second.second
#define all(a)  (a).begin(),(a).end()


signed main(){
  int n,a,b;
  cin>>n>>a>>b;
  V<V<int>> g(n);
  fo(i,n-1){
    g[i].pb(i+1);
    g[i+1].pb(i);
  }
  g[a-1].pb(b-1);
  g[b-1].pb(a-1);
  V<int> ans(n+5,0);

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

    }
  }

  for(int i=1; i<n; i++){
    cout<<ans[i]/2<<endl;
  }
}