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;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef vector<string>vst;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;

#define fastio ios_base::sync_with_stdio(false);cin.tie(0)
#define all(ct) ct.begin() , ct.end()
#define endl "\n"
#define fr(i,a) for(auto i:a)
#define f(i,a,b) for(int i=a;i<b;++i)
#define fd(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define in(d,v) d.find(v)!=d.end()
#define mp make_pair
#define size(a) int(a.size())

const int mod = 1e9 + 7;
const ll inf = ll(1e15);
const double PI = acos(-1);

ll modexp(ll x,ll n,int md){
    if(n==0)return 1;
    if(n%2==0)return modexp((x*x)%md,n/2,md);
    return (x*modexp((x*x)%md,n/2,md))%md;
}


int main(){
    fastio;
    //clock_t tm = clock();
    int n,x,y;
    cin>>n>>x>>y;
    vi ans(n,0);--x;--y;
    f(i,0,n){
        f(j,i+1,n){
            int d=abs(x-i)+abs(y-j)+1;
            ans[min(d,j-i)]++;
        }
    }
    f(i,1,n)cout<<ans[i]<<endl;
    //cout<<((clock()-tm)*1.0)/CLOCKS_PER_SEC<<endl;
}