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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <functional>
#include <vector>
#include <math.h>

using namespace std;
int main()
{
	long long N, X;
	long long x[100010];
	cin >> N >> x[0];

	N = N + 1;

	for (long long i = 1; i < N; i++) {
		cin >> x[i];
	}

	sort(x, x + N);

	for (long long i = 1; i < N; i++) {
		x[i] = x[i] - x[0];
	}
	x[0] = 0;

	long long D = x[1] - x[0];
	long long temp = D;
	for (long long i = 1; i < N; i++) {
		long long a = x[i];
		long long b = D;
		while (temp != 0) {
		   temp = a % b;
		   if (temp == 0) {
		   	D = b;
		   }
		   else {
		   	a = b;
		   	b = temp;
		   }
		}
	}

	cout << D;

	return 0;
}