// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



#define int long long



#define rep(i,n) for(int i=0;i<(n);i++)

#define pb push_back

#define all(v) (v).begin(),(v).end()

#define fi first

#define se second

typedef vector<int>vint;

typedef pair<int,int>pint;

typedef vector<pint>vpint;



template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}

template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}



#define double long double



const int INF=1001001001;

const int INFLL=1001001001001001001ll;

const int mod=1000000007;

inline void add(int &a,int b){

    a+=b;

    if(a>=mod)a-=mod;

}

const double PI=acos(-1);

const double EPS=1e-7;



signed main(){

    int N;cin>>N;

    vint R(N);rep(i,N)cin>>R[i];



    sort(all(R));

    reverse(all(R));

    double ans=0;



    for(int k=3;k<=N;k++){

        vint r(R.begin(),R.begin()+k);

        sort(all(r));

        do{

            vector<double>L(k);

            rep(i,k)L[i]=r[i]*r[(i+1)%k]/2.0;

            double mi=*min_element(all(L));

            double lb=-mi;

            double ub=mi;

            rep(i,100){

                double mid=(ub+lb)/2;

                double sum=0;

                rep(j,k)sum+=acos(mid/L[j]);

                if(sum>=2*PI)lb=mid;

                else ub=mid;

            }



            double sum=0;

            rep(j,k)sum+=acos(lb/L[j]);

            if(abs(sum-2*PI)>EPS)continue;

            double val=0;

            rep(j,k)val+=sin(acos(lb/L[j]))*L[j];

            chmax(ans,val);

        }while(next_permutation(all(r)));

    }

    printf("%.20Lf\n",ans);

    return 0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;



#define int long long



#define rep(i,n) for(int i=0;i<(n);i++)

#define pb push_back

#define all(v) (v).begin(),(v).end()

#define fi first

#define se second

typedef vector<int>vint;

typedef pair<int,int>pint;

typedef vector<pint>vpint;



template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}

template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}



const int INF=1001001001;

const int INFLL=1001001001001001001ll;

const int mod=1000000007;

inline void add(int &a,int b){

    a+=b;

    if(a>=mod)a-=mod;

}

const double PI=acos(-1);

const double EPS=1e-7;



signed main(){

    int N;cin>>N;

    vint R(N);rep(i,N)cin>>R[i];



    sort(all(R));

    reverse(all(R));

    double ans=0;



    for(int k=3;k<=N;k++){

        vint r(R.begin(),R.begin()+k);

        sort(all(r));

        do{

            vector<double>L(k);

            rep(i,k)L[i]=r[i]*r[(i+1)%k]/2.0;

            double mi=*min_element(all(L));

            double lb=-mi;

            double ub=mi;

            rep(i,28){

                double mid=(ub+lb)/2;

                double sum=0;

                rep(j,k)sum+=acos(mid/L[j]);

                if(sum>=2*PI)lb=mid;

                else ub=mid;

            }



            double sum=0;

            rep(j,k)sum+=acos(lb/L[j]);

            if(abs(sum-2*PI)>EPS)continue;

            double val=0;

            rep(j,k)val+=sin(acos(lb/L[j]))*L[j];

            chmax(ans,val);

        }while(next_permutation(all(r)));

    }

    printf("%.20f\n",ans);

    return 0;

}

// Fastest known solution for this problem: