Problem:
Score : 300 points

Problem Statement

A game is played on a strip consisting of N cells consecutively numbered from 1 to N.

Alice has her token on cell A. Borys has his token on a different cell B.

Players take turns, Alice moves first.
The moving player must shift his or her token from its current cell X to the neighboring cell on the left, cell X-1, or on the right, cell X+1.
Note that it's disallowed to move the token outside the strip or to the cell with the other player's token.
In one turn, the token of the moving player must be shifted exactly once.

The player who can't make a move loses, and the other player wins.

Both players want to win. Who wins if they play optimally?

Constraints

2 \leq N \leq 100

1 \leq A < B \leq N

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print Alice if Alice wins, Borys if Borys wins, and Draw if nobody wins.

Sample Input 1

5 2 4

Sample Output 1

Alice

Alice can move her token to cell 3.
After that, Borys will be unable to move his token to cell 3, so he will have to move his token to cell 5.
Then, Alice moves her token to cell 4. Borys can't make a move and loses.

Sample Input 2

2 1 2

Sample Output 2

Borys

Alice can't make the very first move and loses.

Sample Input 3

58 23 42

Sample Output 3

Borys

Solution in JavaScript (node.js v5.12) (judge verdict: Accepted)

function Main(input) {
	input = input.split(" ");
	console.log(((Math.abs(input[1]-input[2])%2)==1)?"Borys":"Alice")
}Main(require("fs").readFileSync("/dev/stdin", "utf8").trim());
function isPrime(n){
    if(false===Number.isInteger(n)){
        return false;
    }
    if(n<=1){
        return false;
    }
    if(2===n){
        return true;
    }
    if(0===n%2){
        return false;
    }
    var square_root = Math.floor(Math.sqrt(n));
    for(var i=3; i<=square_root; i+=2){
        if(0===n%i){
            return false;
        }
    }
    return true;
}
function arr1dto2d (arr){
	for(var i=0;i<arr.length;i++){
		arr[i]=arr[i].split("");
		}
	return arr;
}
function factorialize(num) {
  if (num === 0) { return 1; }
  return num * factorialize(num-1);
}
function arr1dto2d (arr){
	for(var i=0;i<arr.length;i++){
    arr[i]=arr[i].split(" ").map(Number);
    }
  return arr;
}
function rotate(array) {
  const ROW = array.length;
  const COL = array[0].length;
  const row = ROW-1;
  var a = new Array(COL);
  for (var c=0; c<COL; c++) {
    a[c] = new Array(ROW);
    for (var r=0; r<ROW; r++) {
      a[c][r] = array[row-r][c];
    }
  }
  return a;
}
function part_in(base,push,st){
	for(var i=st,p=0;p<push.length;i++,p++){
		base[i] = push[p];
	}
	return base;
}
function part_reverse(arr,s,g){
	var push = arr.slice(s-1,g);
  push.reverse();
	arr = part_in(arr,push,s-1);
	return arr;
}
function zeroPadding(num,length){
    return (num + '000000000000000000000000000000000').slice(0,length);
}
function arrsum(arr){
    var sum = 0;
    for (var i=0,len=arr.length; i<len; ++i) {
        sum += arr[i];
    };
    return sum;
}
function pf(n) {
    var result = {};

    if (n === 1) {
        return {"1":1}
    }

    var init = 2;

    while (n !== 1) {
        var i = init;
        while (i < Number.MAX_SAFE_INTEGER) {
            if (n % i === 0) {
                n /= i;
                if (!(result[i] > 0)) {
                    result[i] = 0;
                }
                result[i]++;
                break;
            }
            i++;
        }
        init = i;
    }
    return result;
}
function gcd(a, b) {
    if (b === 0){
        return a
    }
    return gcd(b, a % b)
}
function lcm(a,b) {
    var g = (n, m) => m ? g(m, n % m) : n
    return a * b / g(a, b)
}
function num(a, b) {
  return a - b;
}
function nibu(a,searchValue){
	var index = -1;

//調べる左端を表す添字(index)
var left = 0;

//調べる右端を表す添字(index)
var right = a.length-1;

//左端と右端にデータがある間は処理を繰り返す
while(left <= right){

    //左右の真ん中を表す添字(index)
    middle =Math.floor((left +right)/2);

    //真ん中の値と探す値を比較する
    if(a[middle]==searchValue){
        //一致した場合、変数に入れて処理終了
        index = middle;
        break;
    }else if(a[middle]<searchValue){
        //探す値より小さい場合、左側にはもっと小さい値しかないので左端の値を真ん中の値の右に移動する
        left = middle +1;
    }else{
        //探す値より大きい場合、右側にはもっと大きい値しかないので右端の値を真ん中の値の左に移動する
        right = middle -1;
    }
}
if(index!=-1) return true;
else return false;
}