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)

from collections import deque
import sys
input = sys.stdin.readline
n,x,y = map(int,input().split())
to = [[] for _ in range(n)]
for i in range(n-1):
    to[i].append(i+1)
    to[i+1].append(i)
x -= 1
y -= 1
to[x].append(y)
to[y].append(x)

ans = [0]*(n-1)
d = {}
Q = deque()
flag = [False]*n
for i in range(n-1):
    Q.clear()
    Q.append((i,-1,0))
    for j in range(n): flag[j] = False
    flag[i] = True
    while Q:
        cur,prev,dist = Q.popleft()
        for nxt in to[cur]:
            if nxt == prev: continue
            if flag[nxt]: continue
            flag[nxt] = True
            if (i,nxt) not in d: ans[dist] += 1
            Q.append((nxt,cur,dist+1))
            d[(i,nxt)] = 1
            d[(nxt,i)] = 1

for i in range(n-1): print(ans[i])