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 Kotlin (1.0.0) (judge verdict: Accepted)

fun max (a:Int,b:Int):Int{
    return if(a>b) a else b
}
fun min (a:Int,b:Int):Int{
    return if(a<b) a else b
}
fun main(args: Array<String>) {
    var (N,X,Y) = readLine()!!.split(" ").map { it.toInt() }
    val c = Y-X +1
    val a = X-1
    val b = N-Y

    for (i in 1 until N) {
        var result = 0
        //円の中
        if(i*2==c)result+=i
        else if(i<=c/2)result+=c
        //円の外から外
        result += max(a+b+2-i,0)
        if(i==1)result-=1
        //円の外から中
        var low=max(-(c/2),1-i)
        var up=min(-1,a-i)
        var t = max(up-low+1,0)*2
        if(c%2==0 && t>0 && c/2<i)t-=1
        if(a+1>=i && i>=2)t-=1
        result+=t

        up=min(-1,b-i)
        t = max(up-low+1,0)*2
        if(c%2==0 && t>0 && c/2<i)t-=1
        if(b+1>=i&& i>=2)t-=1
        result+=t
        println(result)
    }
}