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 Python (3.4.3) (judge verdict: Runtime Error)

import numpy as np
N, X, Y = map(int, input().split())
dist = [0] * (N + 1)
cnt = [0] * (N + 1)

INF = 10**10 + 9

# d = [[INF] * N for _ in range(N)]
d = []
for i in range(N)
    
d = [j - i for i in range(N) for j in range(i, N)]
d = []
for i in range(N):
    tmp = [j - i for j in range(i, N)]
    d.append(tmp)
# d[i][j] = j - i
# d[j][i] = j - i
print('d', d)
exit()
d[X - 1][Y - 1] = 1
d[Y - 1][X - 1] = 1


def warshall_floyd(d):
    #d[i][j]: iからjへの最短距離
    for k in range(N):
        for i in range(N):
            for j in range(N):
                d[i][j] = min(d[i][j], d[i][k] + d[k][j])
    return d


d = warshall_floyd(d)
count = np.bincount(np.array(d).flatten(), minlength=N)
count = count[1:] // 2
print(*count, sep='\n')