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 Pascal (FPC 2.6.2) (judge verdict: Time Limit Exceeded)

const m=1000000000;
var n,x,i,xi,yi,j,min,dd:longint;
a:array[1..100000]of longint;
b:array[1..m] of 0..1;
c:array[1..10000]of 0..1;
begin 
	read(n,x,xi);
	a[1]:=abs(xi-x);
	min:=a[1];
	dd:=2;
	for i:=2 to n do
	begin 
		read(yi);
		if abs(yi-xi)<10000 then 
		if c[abs(yi-xi)]=0 then 
		begin 
			a[i]:=abs(yi-xi);
			c[a[i]]:=1;
			xi:=yi;
			if a[i]<min then min:=a[i];
		end;
	end;
	for i:=1 to n do 
	begin 
		if (i=n) then 
		begin
			while (b[min]=1)or (a[i] mod min<>0) do dec(min);
			write(min);
			exit;
		end;
		for j:=min downto dd do 
			if (b[j]=0) and (a[i] mod j<>0) then b[j]:=1;
		while b[min]=1 do dec(min);
		while b[dd]=1 do inc(dd);
	end;
end.