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 = {}
for i in range(n - 1):
    g.setdefault(i, [])
    g.setdefault(i + 1, [])
    g[i].append(i + 1)
    g[i + 1].append(i)


g[x - 1].append(y - 1)

global_stat = [0] * n

def bfs(i, g):
    q = deque([(i, 0)])
    vis = [0] * len(g)
    vis[i] = 1
    stat = [0] * len(g)
    while q:
        cur_v, cur_dist = q.popleft()
        # print("%s->%s: %s" % (i + 1, cur_v + 1, cur_dist))
        if cur_v > i:
            stat[cur_dist] += 1
        for adj_v in g[cur_v]:
            if not vis[adj_v]:
                vis[adj_v] = 1
                q.append((adj_v, cur_dist + 1))
    # print(stat)
    return stat

for i in range(n):
    local_stat = bfs(i, g)
    for i, v in enumerate(local_stat):
        global_stat[i] += v

# global_stat = [v // 2 for v in global_stat]
# print(global_stat)

for v in global_stat[1:]:
    print(v)