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: Wrong Answer)

def dijkstra(s,n,w,cost):
    #始点sから各頂点への最短距離
    #n:頂点数,　w:辺の数, cost[u][v] : 辺uvのコスト(存在しないときはinf)
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        #まだ使われてない頂点の中から最小の距離のものを探す
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True
               
        for j in range(n):
               d[j] = min(d[j],d[v]+cost[v][j])
    return d

n,x,y=map(int,input().split())
cost = [[float("inf") for i in range(n)] for i in range(n)]
#cost[u][v] : 辺uvのコスト(存在しないときはinf この場合は10**10)
for i in range(1,n):
    cost[i-1][i] = 1
    cost[i][i-1]=1
cost[x-1][y-1]=1
cost[y-1][x-1]=1
ans=[[float("0") for i in range(n)] for i in range(n)] 
for j in range(n):
    ans[j]=dijkstra(j,n,n,cost)
def flatten_2dim(array):
    return [item for sublist in array for item in sublist]
data = flatten_2dim(ans)
for k in range(1,n-1):
    print(int(data.count(k)/2))