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)

object Main extends App {
  import scala.io.StdIn.readLine

  val input = readLine().split(" ").map(_.toInt)
  val N = input(0)
  val X = input(1)

  val xStr = readLine()
  val xArray = (X + " " + xStr).split(" ").map(_.toInt).sorted

  var min1 = 10000000000L
  var min2 = 10000000000L
  var flag = false
  for (i <- 0 to N - 1) {
    val diff = xArray(i + 1) - xArray(i)
    if (min1 > diff) min1 = diff
    if (i == 0) {
      min2 = diff
    } else {
      if (min1 == min2 && min2 < diff) {
        min2 = diff
      } else if (min1 < diff && diff < min2) {
        min2 = diff
      }
    }
  }

  if (min2 % min1 == 0) {
    print(min1)
  } else {
    var res = 1L;
    for (i <- 2L to min1 / 2L) {
      if (min1 % i == 0 && min2 % i == 0) res = i
    }
    print(res)
  }

}