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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n, X; scanf("%d%d", &n, &X);
    int x;
    vector<int> diff;
    for (int i = 0; i < n; i++) {
        scanf("%d", &x);
        diff.push_back(abs(X-x));
    }
    sort(diff.begin(), diff.end());
    unique(diff.begin(), diff.end());

    vector<int> dev;
    for (int i = 1; i <= pow(diff[0], 0.5); i++) {
        if (diff[0]%i==0) {
            dev.push_back(i);
            dev.push_back(diff[0]/i);
        }
    }
    sort(dev.begin(), dev.end());
    unique(dev.begin(), dev.end());
    reverse(dev.begin(), dev.end());

    int num = 0;
    int ret = dev[num];
    for (int i = 1; i < diff.size(); i++) {
        while (diff[i] % dev[num] != 0) {num++;}
        ret = dev[num];
        if (ret == 1) break;
    }

    printf("%d\n", ret);
    return 0;
}