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: Accepted)

def examA():
    S = SI()
    if S[2]==S[3] and S[4]==S[5]:
        print("Yes")
    else:
        print("No")
    return

def examB():
    X = I()
    ans = 0
    ans += (X//500) * 1000
    X %= 500
    ans += (X//5) * 5
    print(ans)
    return

def examC():
    K, N = LI()
    A = LI()
    L = [0]*N
    L[0] = K-A[-1] + A[0]
    for i in range(1,N):
        L[i] = A[i]-A[i-1]
    ans = sum(L) - max(L)
    #print(L)
    print(ans)
    return

def examD():
    def bfs(n, e, fordfs):
        # 点の数、スタートの点、有向グラフ
        W = [-1] * n
        # 各点の状態量、最短距離とか,見たかどうかとか
        W[e] = 0
        que = deque()
        que.append(e)
        while que:
            now = que.popleft()
            nowW = W[now]
            for ne in fordfs[now]:
                if W[ne] == -1:
                    W[ne] = nowW + 1
                    que.append(ne)
        return W
    N, X, Y = LI()
    V = [[]for _ in range(N)]
    for i in range(N-1):
        V[i].append(i+1)
        V[i+1].append(i)
    V[X-1].append(Y-1)
    V[Y-1].append(X-1)
    #print(V)
    ans = [0]*(N-1)
    for i in range(N):
        L = bfs(N,i,V)
        for l in L:
            if l==0:
                continue
            ans[l-1] += 1
    for v in ans:
        print(v//2)
    return

def examE():
    ans = 0
    print(ans)
    return

def examF():
    ans = 0
    print(ans)
    return

import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]

sys.setrecursionlimit(10**6)

if __name__ == '__main__':
    examD()

"""

"""