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 <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define Cout(s) cout << s << endl;
#define CoutV(vec)        \
    for (auto v : vec)    \
    {                     \
        cout << v << " "; \
    }                     \
    cout << endl;
#define CoutVV(vecvec)      \
    for (auto vec : vecvec) \
        CoutV(vec);

using ll = long long;
using P = pair<int, int>;
#define INF 10000000;

ll MOD = 1e9 + 7;

ll gcd(ll x, ll y) { return (y != 0) ? gcd(y, x % y) : x; }

int main()
{


    ll n, x;
    cin >> n >> x;
    vector<ll> X(n);
    vector<ll> d(n - 1);
    REP(i, n)
    {
        cin >> X[i];
    }
    if (n == 1)
    {
        Cout(abs(x - X[0]));
        return 0;
    }

    ll ans = 0;
    REP(i, n - 1)
    d[i] = abs(X[i] - X[i + 1]);
    REP(i, n - 1)
    ans = max(ans, gcd(d[i + 1], d[i]));

    cout << ans << endl;
    return 0;
}