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: Runtime Error)

from sys import stdin
from collections import deque


class BreadthFirstSearch:
    def __init__(self, num_nodes):
        self.n = num_nodes
        self.graph = [[] for _ in range(self.n)]

    # Edge数分回すことを想定
    def add_edge(self, start, end, undirected=True):
        self.graph[start].append(end)
        if undirected:
            self.graph[end].append(start)

    # depth = distance
    def search(self, start_node):
        order = [-1] * self.n # a bfs ordering of each vertex
        parent = [-1] * self.n # parent of each vertex in the bfs search tree
        depth = [-1] * self.n # the depth of each vertex
        q = deque([(start_node, -1, 0)]) # (vertex, parent, depth)
        num = 0 # current ordering
        while len(q) > 0:
            v, p, d = q.popleft()
            if order[v] < 0: # visited v for the first time
                order[v] = num; parent[v] = p; depth[v] = d
                num += 1
                for u in adj_list[v]:
                    if order[u] >= 0: continue
                    q.append((u, v, d+1))
        return order, parent, depth



# BFS
def breadth_first_search(adj_list, start_node):
    N = len(adj_list)
    order = [-1] * N # a bfs ordering of each vertex
    parent = [-1] * N # parent of each vertex in the bfs search tree
    depth = [-1] * N # the depth of each vertex
    q = deque([(start_node, -1, 0)]) # (vertex, parent, depth)
    num = 0 # current ordering
    while len(q) > 0:
        v, p, d = q.popleft()
        if order[v] < 0: # visited v for the first time
            order[v] = num; parent[v] = p; depth[v] = d
            num += 1
            for u in adj_list[v]:
                if order[u] >= 0: continue
                q.append((u, v, d+1))
    return order, parent, depth


# BFS pypyでAC
def get_result(data):
    N, X, Y = data
    _data = [[i, i+1] for i in range(1, N)]
    _data += [[X, Y]]
    bfs = BreadthFirstSearch(N)
    for val in _data:
        bfs.add_edge(val[0], val[1])
    dist_cnt = [0 for i in range(N)]
    for i in range(N):
        _, _, depth = bfs.search(i)
        for j in range(i, N):
            dist_cnt[depth[j]] += 1
    for i in range(1, N):
        print(dist_cnt[i])


if __name__ == '__main__':
    data = list(map(int, stdin.readline().rstrip('\n').split(' ')))
    get_result(data)