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>
using namespace std;
typedef long long int ll;
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#define  boost ios_base::sync_with_stdio(false);cin.tie(NULL);
#define ull unsigned long long
#define d1(x) cout<<#x<<" "<<x<<endl;
#define d2(x,y) cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<endl;
#define d2i(x,y,i) cout<<#x<<i<<" "<<x<<" "<<#y<<i<<" "<<y<<endl;

#define fr(i,l,r) for(ll i=l;i<r;i++)
#define mems(a,x) memset(a,x,sizeof(a))
#define mod 1000000007
#define ff first
#define ss second
#define pb(x) push_back(x)
#define vll vector<ll>
#define pbp(x,y) push_back(make_pair(x,y))
#define  mat vector<vector<ll>>
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

void solve(){
ll n,a,b;
cin>>n>>a>>b;
vll d(n);
vll v(n+1);
fr(i,1,n)
v[i]=i;


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

          }
}
fr(i,1,n){
    cout<<d[i]<<endl;
}



}

int main(){
    ll t=1;
  /// cin>>t;
    while(t--){
        solve();
        cout<<endl;
    }
    return 0;
}