#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0;i<((int)(n));i++)
#define reg(i,a,b) for(int i = ((int)(a));i<=((int)(b));i++)
#define irep(i,n) for(int i = ((int)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(int i = ((int)(b));i>=((int)(a));i--)
typedef long long ll;

/*
*/

ll n,a[100010],b[100010],ru[100010],ru2[100010],ans=1e18;
set<ll> s;

void init(){
	cin>>n;
	reg(i,1,n)cin>>a[i];
	sort(a+1,a+n+1);
	reg(i,1,n)s.insert(a[i]);
	ru[0]=0;
	reg(i,1,n)ru[i]=ru[i-1]+a[i];
	ireg(i,1,n)b[i]=b[n]-a[i];
	ru2[n+1]=0;
	ireg(i,1,n)ru2[i]=ru2[i+1]+b[i];
}

int main(void){
	init();
	if(s.size()==1){
		ans=1;
	}else{
		reg(i,1,n-1){  // 1~iまでを同じ数字にする
			ll sum1=0,sum2=0,sum3=0,sum4=0;
			sum1 += ru2[1]-ru2[(i+1)/2+1]-((i+1)/2-1+1)*b[(i+1)/2];
			sum2 += ru[i]-ru[(i+1)/2-1]-(i-(i+1)/2+1)*a[(i+1)/2];
			sum3 += ru2[i+1]-ru2[(n+i+1)/2+1]-((n+i+1)/2-(i+1)+1)*b[(n+i+1)/2];
			sum4 += ru[n]-ru[(n+i+1)/2-1]-(n-(n+i+1)/2+1)*a[(n+i+1)/2];
			// cout<<sum1<<" "<<sum2<<" "<<sum3<<" "<<sum4<<endl;
			ans=min(ans, sum1+sum2+sum3+sum4);
		}
	}
	cout<<ans<<endl;
	return 0;
}