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 <math.h> 
#include <algorithm>
#include <numeric>
#include <functional>
#include <vector>
#include <string>
#include <numeric>
#include <bitset>
#include <queue>

using namespace std;
vector<long long> pt;
vector<long long> ans;


int main() {
	//input
	long long n,x;
	cin >> n >> x;

	for (long long i = 0; i < n; ++i) {
		long long tmp = 0;
		cin >> tmp;
		pt.push_back(tmp);
	}

	long long cnt_kisu = 0;
	long long cnt_gusu = 0;

	int cnt = 0;
	if (n != 1) {
		for (long long t = 0; t < pt.size(); ++t) {
			if (x == pt.at(t)) {
				cnt++;
			}
		}
		if (cnt == 0) {
			pt.push_back(x);
		}
		sort(pt.begin(), pt.end());
		//全ての値の差が偶数の時
		for (long long s = 0; s < pt.size() - 1; ++s) {
			if ((pt.at(s + 1) - pt.at(s)) % 2 != 0) {
				//奇数だったらインクリメント
				cnt_kisu++;
			}
			else {
				cnt_gusu++;
			}
		}
		if (cnt_kisu == pt.size() - 1 || cnt_gusu == pt.size() - 1) {
			//隣あう値で一番小さい値を求める
			for (long long k = 0; k < pt.size() - 1; ++k) {
				long long atmp = 0;
				long long a = pt.at(k + 1);
				long long b = pt.at(k);
				atmp = pt.at(k + 1) - pt.at(k);
				ans.push_back(atmp);
			}
			sort(ans.begin(), ans.end());
			cout << ans.at(0) << endl;
			return 0;
		}
		else {
			cout << 1 << endl;
			return 0;
		}

	}
	else {
		cout << pt.at(0) - x << endl;
	}

	return 0;
}