#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>

using namespace __gnu_pbds;
using namespace std;

// using long doubles saves you from corner cases but is very time consuming
#define double      	long double
#define int         	long long
#define pb          	push_back
#define pii         	pair<int,int>
#define vi          	vector<int>
#define vii         	vector<pii>
#define mi          	map<int,int>
#define mii         	map<pii,int>
#define all(a)      	(a).begin(),(a).end()
#define sz(x)       	(int)x.size()
// just comment the line below in case of interactive problems
#define endl        	"\n"
#define repp(i,a,b) 	for(int i=a;i<b;i++)
#define rep(i,a,b) 		for(int i=a;i<=b;i++)
#define brep(i,a,b)     for(int i=a;i>=b;i--)
#define deb1(x)      	cout << #x << "=" << x << endl
#define deb2(x, y)  	cout << #x << "=" << x << "," << #y << "=" << y << endl
#define deb3(x, y, z)  	cout << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define trace(v) 		cout << #v << "=";for(auto it=v.begin();it!=v.end();it++)cout<<*it<<" ";cout<<endl;
#define tracearr(a,l,r) cout << #a << "=";for(int iii=l;iii<=r;iii++)cout << a[iii] << " ";cout << endl;
#define PI          	3.1415926535897932384626
#define F 				first
#define S 				second
#define clr(x,y) 		memset(x, y, sizeof(x))
#define IOS         	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//vector<vector<int> > v( n , vector<int> (m, 0));


typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
// *s.find_by_order(1) - gives the 2nd samllest element in set
// s.order_of_key(x)   - gives the number of elements in the set which are strictly smaller than x


const int N=2000 + 5;
const int MOD=1e9 + 7;

int n;
int dp[N][N];
vi v;

int solve(int i,int j){
    if(i==-1 && j==n)
        return 0;
    int& ans = dp[i][j];
    if(ans!=-1)
        return ans;
    ans = 1e18;
    deb2(i,j);
    if(i==-1)
        ans = v[j] - v[0] + solve(i,j+1);
    else if(j==n)
        ans = v[n-1] - v[i] + solve(i-1,j);
    else{
        ans = min(v[j]-v[i+1] + solve(i,j+1),v[j-1] - v[i] + solve(i-1,j));
    }
    return ans;
}

void test_case()
{
    cin >> n;
    rep(i,1,n){
        int x;
        cin >> x;
        v.pb(x);
    }
    sort(all(v));
    if(n==1){
        cout << "0" << endl;
        return;
    }
    else if(n==2){
        cout << v[1] - v[0] << endl;
        return;
    }
    clr(dp,-1);
    int mid = n/2;
    int ans = solve(mid-1,mid+1);
    cout << ans << endl;
}

int32_t main()
{
	//uncomment the below 2 lines and for simplicity use t=1,when stress testing,
	//freopen("input1.txt", "r", stdin);
	//freopen("original_output.txt", "w", stdout);
	IOS;
	int T=1;
	//cin >> T;
	// int numOfSetBits = __builtin_popcountll(n);
	std::cout << std::fixed << std::setprecision(12);
	while(T--)
	{
		test_case();
	}
}

/*
	* while writing bool comp function for sorting put 1 statement in the end which has no if comdition
 	* in case of multiple test cases and N=1e5 don't intialize arrays as global except for graph vector-array
 	* always check whether or not you are doing mod of a negative number
 	* always use 1LL instead of 1
*/