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 <cstdio>
#include <vector>
#include <map>
#include <cmath>
#include <string>
#include <algorithm>

#define REP(i, n) for(int i = 0; i != n; ++i)
#define REPR(i, n) for(int i = n - 1; i != -1; --i)
#define FOR(i, a, b) for(int i = a; i != b; ++i)
#define IN(n) (cin >> n)
#define OUT(n) (cout << n << endl)

using namespace std;
using ll = unsigned long long;

int main() {
	int N, X;
	IN(N);
	IN(X);
	vector<int> x(N), d(N);
	int i, j;
	for(i = 0; i < N; i++) {
		IN(x[i]);
		d[i] = abs(x[i] - X);
	}
	
	sort(d.begin(), d.end());
	
	int t = 1;
	
	vector<int> memo;
	
	bool f = false;
	
	auto itr = memo.begin();
	
	for(i = d[0]; i > 1; i--) {
		for(j = 0; j < N; j++) {
			if(d[j] % i != 0) {
				goto A;
			}
		}
		t = i;
	}
	A:
	cout << t << endl;
}