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: Time Limit Exceeded)

fun main(args: Array<String>) {
    val (N, X, Y) = readLine()!!.split(" ").map(String::toInt)

    // グラフ作成
    val G = Array(N) { mutableListOf<Int>() }
    for (i in 0 until N) {
        if (i != N - 1) {
            G[i].add(i + 1)
        }
        if (i != 0) {
            G[i].add(i - 1)
        }
    }
    G[X - 1].add(Y - 1)
    G[Y - 1].add(X - 1)


    // 探索処理
    var keiro = mutableListOf<Int>()
    fun dfs(d: Int, v: Int): Set<Int> {
        if (d == 1) {
            return G[v].filterNot { keiro.contains(it) }.toSet()
        }

        val set = mutableSetOf<Int>()
        for (_v in G[v]) {
            keiro.add(_v)
            set.addAll(dfs(d - 1, _v))
        }
        return set

    }

    for (i in 1 until N) {
        var ans = mutableSetOf<Pair<Int, Int>>()
        for (v in 0 until N) {
            // 通った経路
            keiro = mutableListOf(v)
            val tmp = dfs(i, v)
            // 通った経路は距離iではないので除外する
            // i to j i < j にして重複を取り除く
            ans.addAll(tmp.filterNot { keiro.contains(it) }
                .map {
                    if (v < it) {
                        v to it
                    } else {
                        it to v
                    }
                }
            )
        }
        println(ans.size)
    }
}