Problem:
Score : 300 points

Problem Statement

There are N cities on a number line. The i-th city is located at coordinate x_i.

Your objective is to visit all these cities at least once.

In order to do so, you will first set a positive integer D.

Then, you will depart from coordinate X and perform Move 1 and Move 2 below, as many times as you like:

Move 1: travel from coordinate y to coordinate y + D.

Move 2: travel from coordinate y to coordinate y - D.

Find the maximum value of D that enables you to visit all the cities.

Here, to visit a city is to travel to the coordinate where that city is located.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq X \leq 10^9

1 \leq x_i \leq 10^9

x_i are all different.

x_1, x_2, ..., x_N \neq X

Input

Input is given from Standard Input in the following format:

N X
x_1 x_2 ... x_N

Output

Print the maximum value of D that enables you to visit all the cities.

Sample Input 1

3 3
1 7 11

Sample Output 1

2

Setting D = 2 enables you to visit all the cities as follows, and this is the maximum value of such D.

Perform Move 2 to travel to coordinate 1.

Perform Move 1 to travel to coordinate 3.

Perform Move 1 to travel to coordinate 5.

Perform Move 1 to travel to coordinate 7.

Perform Move 1 to travel to coordinate 9.

Perform Move 1 to travel to coordinate 11.

Sample Input 2

3 81
33 105 57

Sample Output 2

24

Sample Input 3

1 1
1000000000

Sample Output 3

999999999

Solution in Scala (2.11.7) (judge verdict: Wrong Answer)

import java.io.{InputStream, OutputStream, PrintStream}
import java.util.Scanner

object Main {

  def main(args: Array[String]): Unit = {
    val solver = new Solver(System.in, System.out)
    solver.solve()
  }

  class Solver(in: InputStream, out: OutputStream) {
    lazy val input = new Scanner(in)

    lazy val output = new PrintStream(out)

    def solve(): Unit = {
      val Array(n, x) = input.nextLine().split(" ").map(_.toInt)
      val xs          = input.nextLine().split(" ").map(_.toInt)

      val diffs   = xs.map(xi => Math.abs(xi - x))
      val minDiff = diffs.min
      val divisors = minDiff :: 1.to(Math.sqrt(minDiff).toInt).filter(i => minDiff % i == 0).reverse.toList

      divisors.find(d => diffs.forall(_ % d == 0)) match {
        case Some(v) => println(v)
      }
    }

  }

}