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)

from collections import deque
n, x, y = map(int, input().split())
g = dict()
for i in range(n):
    if i == 0:
        g[i] = [i+1]
    elif i == n-1:
        g[i] = [i-1]
    else:
        g[i] = [i-1, i+1]
g[x-1].append(y-1)
g[y-1].append(x-1)
# print(g)
distance = [0]*(n-1)
# 1番ノードから各ノードへの最短距離を求めていく.
for i in range(n):
    queue = deque()
    visited = [0]*n
    # まず処理される予定のノードが格納されているqueueに最初のノードを格納する
    queue.append(i)
    while queue:
        target = queue.popleft()
        for j in g[target]:
                if visited[j] == 0 or visited[j] > visited[target] + 1:
                    visited[j] = visited[target] + 1
                    queue.append(j)
    for k in range(len(visited)):
        if k > i:
            if visited[k] >= 1:
                distance[visited[k]-1] += 1
    # print(f"{i}番ノードからの各ノードへの最短距離は{visited}である")
for l in distance:
    print(l)