3579

/*
10万个数，所有的两两相减的绝对值的中数，即中间的数

*/

// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef __int64 LL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)

#define Z(a,b) ((a)<<(b))
#define Y(a,b) ((a)>>(b))

const double eps = 1e-6;
const double INFf = 1e100;
const int INFi = 1000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}


// code begin
// 两次二分，一次二分答案，一次二分位置
//
int N;
int data[100010];
int main()
{
	read;
	write;
	while(scanf("%d",&N)==1)
	{
		for(int i=0;i<N;i++)
		{
			scanf("%d",data+i);
		}
		sort(data,data+N);

		LL pos = (N*(N-1)/2)&1?(N*(N-1)/4+1):(N*(N-1)/4);
		int l = 0,r = 1000000001,mid,leftpos,leftpos2,ll,rr,midd,ans = -1,flag;
		while(l<r)
		{
			mid = (l+r)>>1;
			leftpos = 0;
			flag = 0;
			for(int i=0;i<N;i++)
			{
				ll = i+1;rr = N;
				while(ll<rr)
				{
					midd = (ll+rr)>>1;
					if(data[midd]-data[i]==mid)
					{
						flag = 1;
					}
					if(data[midd]-data[i]>mid)
					{
						rr = midd;
					}
					else
					{
						ll = midd+1;
					}
				}
				leftpos += rr-i-1;
			}
			if(flag && pos==leftpos)
			{
				ans = mid;
				break;
			}
			if(leftpos>=pos)
			{
				ans = mid;
				r = mid;
			}
			else
			{
				l = mid+1;
			}
		}

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