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 Python (3.4.3) (judge verdict: Accepted)

N, X, Y = map(int, input().split())
graph = [[i + 1] for i in range(N - 1)]
graph.append([])
for i in range(1, N):
    graph[i].append(i-1)
graph[X-1].append(Y-1)
graph[Y-1].append(X-1)

# print(graph)
from collections import deque
def bfs_graph(G, start):
    # 隣接リストを食わせる
    dist = [-1] * N
    que = deque([start])
    dist[start] = 0
    while que:
        v = que.popleft()
        d = dist[v]
        for w in G[v]:
            if dist[w] > -1:
                continue
            dist[w] = d + 1
            que.append(w)
    return dist

ans = [0] * (N - 1)
for s in range(N - 1):
    res = bfs_graph(graph, s)
    # print(res)
    for val in res[s+1:]:
        if val >= 1:
            ans[val-1] += 1
print(*ans, sep="\n")