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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys

from collections import Counter, defaultdict, deque

input = sys.stdin.readline

N, X, Y = [int(x) for x in input().split()]


def bfs(source):
    q = deque([source])
    dist = {source: 0}
    while q:
        i = q.popleft()
        d = dist[i]
        nbr = []
        if i - 1 >= 1:
            nbr.append(i - 1)
        if i + 1 <= N:
            nbr.append(i + 1)
        if i == X:
            nbr.append(Y)
        if i == Y:
            nbr.append(X)
        for j in nbr:
            if j not in dist:
                q.append(j)
                dist[j] = d + 1
    return dist


count = Counter()
for i in range(1, N + 1):
    dist = bfs(i)
    assert len(dist) == N
    for j, d in dist.items():
        if i < j:
            count[d] += 1

ans = []
for k in range(1, N):
    ans.append(str(count[k]))
print('\n'.join(ans))