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 ll;
#define FAST_IO ios_base::sync_with_stdio(0); cin.tie(0);
//#define cerr if(false)cerr
#define PRINT(x) cerr<<"\n" << (#x) << " is " << (x) << endl;
//Erase duplicates in vector
#define REMOVE_DUP(x) x.erase(unique(x.begin(), x.end()), x.end());
#define UP toupper // lowercase to uppercase;
#define DOWN tolower
#define SZ(a) (int)a.size()
#define EB emplace_back
#define PB push_back
#define REP(i, a, b) for(int i=a; i<b; i++)
#define PER(i, a, b) for(int i=a; i>=b; i--)
ostream_iterator<int> output(cout, " ");
#define MOD 1e9+7
#define INF INT_MAX;
#define lim_size 200007


void solve(){
	int n, x, y;
	cin >> n >> x >> y;
	unordered_map<int, int> m;
	for(int i=1; i<n; i++){
		for(int j=i+1; j<=n; j++){
			int dist = min(j - i, abs(i-x) + 1 + abs(j-y));
			m[dist]++;
		}
	}
	for(int i=1; i<n; i++){
		cout << m[i] << endl;
	}
}

int main(){
	FAST_IO;
	int t=1; //cin >> t;
	while(t--){
		solve();
	}
}