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 (7.3.0) (judge verdict: Accepted)

import bisect, collections, copy, heapq, itertools, math, string
import sys
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int, sys.stdin.readline().rstrip().split())
def LI(): return list(map(int, sys.stdin.readline().rstrip().split()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())

from collections import deque
def bfs(u, n, g):
    queue = deque([u])
    d = [None] * n # uからの距離の初期化
    d[u] = 0 # 自分との距離は0
    while queue:
        v = queue.popleft()
        for i in g[v]:
            if d[i] is None:
                d[i] = d[v] + 1
                queue.append(i)
    return d

N, X, Y =MI()
E = [[] for _ in range(N + 1)]
for i in range(1, N):
    a = i
    b = i + 1
    E[a].append(b)
    E[b].append(a)
E[X].append(Y)
E[Y].append(X)

dist = [0] * N

for i in range(1, N + 1):
    D = deque(bfs(i, N + 1, E))
    D.popleft()
    for d in D:
        dist[d] += 1

for i in range(1, N):
    print(dist[i] // 2)