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)

#!usr/bin/env pypy3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect


def LI(): return [int(x) for x in sys.stdin.readline().split()]


def I(): return int(sys.stdin.readline())


def LS(): return [list(x) for x in sys.stdin.readline().split()]


def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res


def IR(n):
    return [I() for i in range(n)]


def LIR(n):
    return [LI() for i in range(n)]


def SR(n):
    return [S() for i in range(n)]


def LSR(n):
    return [LS() for i in range(n)]


sys.setrecursionlimit(1000000)


def main():
    N, X, Y = LI()
    X -= 1
    Y -= 1
    dist = [[N] * N for _ in range(N)]

    q = deque()
    for i in range(N):
        q.append(i)
        dist[i][i] = 0
        while q:
            u = q.popleft()
            if u < N - 1:
                if dist[i][u + 1] > dist[i][u] + 1:
                    q.append(u + 1)
                    dist[i][u + 1] = dist[i][u] + 1
            if u >= 1:
                if dist[i][u - 1] > dist[i][u] + 1:
                    q.append(u - 1)
                    dist[i][u - 1] = dist[i][u] + 1
            if u == X:
                if dist[i][Y] > dist[i][u] + 1:
                    q.append(Y)
                    dist[i][Y] = dist[i][u] + 1
            if u == Y:
                if dist[i][X] > dist[i][u] + 1:
                    q.append(X)
                    dist[i][X] = dist[i][u] + 1

    cnt = [0] * N

    for i in range(N):
        for j in range(i + 1, N):
            cnt[dist[i][j]] += 1

    for i in cnt[1:]:
        print(i)


main()