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)

//xenowi`tz -- Jai Shree Ram
#include<bits/stdc++.h>
using namespace std;

#define fori(i,a,b) for (long long int i = a; i <= b ; ++i)
#define ford(i,a,b) for(long long int i = a;i >= b ; --i)
#define mk make_pair
#define mod 1000000007
#define MAXN 400005
#define pb push_back
#define vec vector<long long int>
#define ll long long
#define rnd mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count())
#define pi pair<long long int,long long int>
#define sc second
#define fs first

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int n, x, y;
	cin >> n >> x >> y;
	vector<int> arr(400000,0);
	for (int i = 1; i <= n; i++) {
		for (int j = i+1; j <= n; j++) {
			int dist = min(abs(i-x)+abs(j-y)+1, j-i);
			arr[dist] += 1;
		}
	}
	for (int k = 1; k < n; k++) {
		cout << arr[k] << " ";
	}
	cout << endl;
	return 0;
}