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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program skip
  implicit none
  integer :: n, x(100001) = 0, i, m
  read(*,*) n, x(1)
  read(*,*) x(2:n+1)
  call quick_sort(x(1:n+1))
  m = x(2)-x(1)
  do i = 2, n
    m = gcd(m,x(i+1)-x(i))
  end do
  write(*,'(i0)') m
contains
  integer function gcd(a,b)
    integer, intent(in) :: a, b
    integer :: x, y
    x = max(abs(a),abs(b))
    y = min(abs(a),abs(b))
    do while (y > 0)
      gcd = y
      y = mod(x,y)
      x = gcd
    end do
  end
  recursive subroutine quick_sort(a)
    integer, intent(inout) :: a(:)
    integer :: n, l, r, m
    integer :: p
    n = size(a)
    l = 1
    r = n
    m = (n+1)/2
    p = a(l)+a(m)+a(r)-max(a(l),a(m),a(r))-min(a(l),a(m),a(r))
    do
      do while (a(l) < p)
        l = l+1
      end do
      do while (a(r) > p)
        r = r-1
      end do
      if (l >= r) exit
      a(l) = xor(a(l),a(r))
      a(r) = xor(a(l),a(r))
      a(l) = xor(a(l),a(r))
      l = l+1
      r = r-1
    end do
    if (l-1 > 1) call quick_sort(a(1:l-1))
    if (n > r+1) call quick_sort(a(r+1:n))
  end
end program skip