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)

///*****************************************************///
///*.......................Faiyaj Bin Amin............................*///
///*...................................................................*///
///*.........................KUET CSE 2K17......................*///
///*...................................................*///
///*****************************************************///

#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

#define fastio      ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)
#define ret return
#define rep(q, i, n) for(int q=i; q<n; q++)
#define forn(i, n) for(int i=0; i<n; i++)
#define mems(a,x) memset(a,x,sizeof(a))
#define ll long long
#define pb push_back
#define em emplace_back
#define mp make_pair
#define all(v) (v).begin(), (v).end()
#define ff first
#define ss second
#define pii pair<int,int>
#define pdd pair<double, double>

#define see(x)        cerr<< __LINE__ << ": " << (#x) << " is "<< (x) <<'\n';
#define seecontent(v) for(auto x : v) { cout << x << " ";} cout << "\n\n";

#define PI 2*acos(0.0)
#define lcm(a, b)  ((a / __gcd(a, b) ) * b)

#define filein freopen ("input.txt", "r", stdin)
#define fileout freopen ("output.txt", "w", stdout)

#define mx 20005
#define INF 1e18
#define MOD 1000000007
#define MIN 0

//template <typename T>  using orderedSet =
//    tree<T, null_type, less_equal<T>,
//    rb_tree_tag, tree_order_statistics_node_update>; //find_by_order, order_of_key


/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0}; // Directional
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

/**------------------------------------BITMASK------------------------------------------**/
//int Change(int N, int pos){return N= N^(1<<pos);}
//bool Check(int N, int pos){return (bool)(N &(1<<pos));}  /// check bit 0 or 1
/**-----------------------------------------------------------------------------------------------**/

//using namespace __gnu_pbds;
using namespace std;

bool vis[mx+5];
int dis[mx+5];
int ks[mx+1]= {0};
vector<int> adj[mx+3];

void BFS(int src, int n)
{

////    see(src);
    memset(vis,0, sizeof vis);

    for(int i=0; i<=n; i++)
        dis[i]=0;

    queue <int> q;
    vis[src]=1;
    q.push(src);

    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int v:adj[u])
        {
            if(!vis[v])
            {
                vis[v]=1;
                dis[v]=dis[u]+1;

////                see(u);
////                see(v);
////                see(dis[v]);
                ks[dis[v]]++;
                q.push(v);
            }
        }
    }
////    cout << endl;


}

int main()
{
    ///FASTIO;

    int n,x,y;
    cin>>n>>x>>y;

    for(int i=1; i<n; i++)
    {
        adj[i].push_back(i+1);
        adj[i+1].push_back(i);
    }
    adj[x].push_back(y);
    adj[y].push_back(x);


    for(int i=1; i<=n; i++)
        BFS(i,n);

////    for(int i=1; i<=n; i++)
////    {
////        cout << i << " " << dist[i]) << endl;
////    }
    for(int i=1; i<n; i++)
    {
        cout << ks[i]/2 << "\n";
    }

    return 0;
}