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: Time Limit Exceeded)

import sys
sys.setrecursionlimit(10**9)
N,X,Y = map(int,input().split())
G = dict()
for i in range(0,N):
    G[i] = []
for i in range(0,N-1):
    G[i].append(i+1)
    G[i+1].append(i)
G[X-1].append(Y-1)
G[Y-1].append(X-1)
ans = [0 for _ in range(N)]
def bfs(v,G,queue,dist):
    seen[v] = True
    for vertex in G[v]:
        if seen[vertex]:
            continue
        elif vertex in queue:
            continue
        queue.append(vertex)
        dist[vertex] = dist[v] + 1
    for elements in queue:
        queue.remove(elements)
        bfs(elements,G,queue,dist)
for k in range(0,N):
    dist = [-1 for _ in range(N)]
    dist[k] = 0
    seen = [False for _ in range(N)]
    queue = []
    bfs(k,G,queue,dist)
    for j in range(1,N):
        a = dist.count(j)
        ans[j-1] += a
for j in range(1,N):
    print(ans[j-1]//2)