#include <fstream>
#include <algorithm>
using namespace std;
ifstream fi("marmot.in");
ofstream fo("marmot.out");
const int NMAX=1e6+5,INF=1e9;
int n,s[NMAX],sum;
pair <int,int> m[NMAX],in[NMAX];
void structure(int nr)
{
    if(in[-nr].first<0)
    {
        fo<<"{";
        structure(in[-nr].first);
        fo<<"}";
    }
    else fo<<"("<<s[in[-nr].first]<<")";
    if(nr==-n+1)
        fo<<"-root-";
    else fo<<"-i-";
    if(in[-nr].second<0)
    {
        fo<<"{";
        structure(in[-nr].second);
        fo<<"}";
    }
    else fo<<"("<<s[in[-nr].second]<<")";
}
int main()
{
    ///n=number of marmots
    fi>>n;
    for(int i=1;i<=n;i++)
    {
        fi>>m[i].first;///m[i].first=number of awakenings for marmot i
        m[i].second=i;///m[i].second=the indice before sorting of the marmot i
        ///(we need this number for the structure)
        s[i]=m[i].first;
    }
    sort(m+1,m+n+1);
    for(int i=1;i<n;i++)
    {
        sum+=m[1].first+m[2].first;
        m[1].first=m[1].first+m[2].first; /// we give m[1] the value of the new intersection created
        in[i]={m[1].second,m[2].second};
        m[1].second=-i;
        m[2].first=INF;///we give m[2] a big value in order to avoid using it again
        sort(m+1,m+n+1);
    }///we repeat this process n-1 times, because in the end we need to have one intersection left, which is the root
    fo<<"minimal sum is "<<sum<<"\n";///sum=answer to our problem, the sum of weights
    ///In this algorithm we find the minimal sum in complexity O(N^2*log2N)
    structure(-(n-1));
}