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 PyPy2 (5.6.0) (judge verdict: Runtime Error)

# 入力
N, X, Y = map(int, input().split())
Graph = [[] for _ in range(N)]
for i in range(N):
    if i < N - 1:
        Graph[i].append(i + 1)
        Graph[i + 1].append(i)
    if X == i:
        Graph[X - 1].append(Y - 1)
        Graph[Y - 1].append(X - 1)
# print(Graph)
ans = [0] * (N - 1)
from collections import deque
# 全ての点からDFSを行い、最短距離を求める
for i in range(N):
    dist = [-1]*N
    que = deque([i])
    dist[i] = 0
    while que:
        v = que.popleft()
        d = dist[v]
        for w in Graph[v]:
            if dist[w] > -1:
                continue
            dist[w] = d + 1
            que.append(w)
    # print(dist)
    # 最短距離の個数をansに格納する
    for d in dist:
        if d > 0:
            ans[d - 1] += 1
# 全ての点からの最短距離を格納した。
# 条件1 <= i < j <= Nより、1 => 2はカウントするが、2 => 1はカウントしない。
# そのため、2で割って帳尻合わせを行う。
for a in ans:
    print(a // 2)