#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
#define maxn 100
struct node
{
    int v,f,ls,rs;
}tree[4*maxn];
int num;//num表示總節點個數
int judge()
{
    int sum=0;
    for(int i=0;i<num;i++)
    {
        if(tree[i].f==-1)
            sum++;
        if(sum==2) return 1;
    }
    return 0;
}
void bfs(int*p)
{
    queue<struct node>q;
    int i=0;
    while(!q.empty()) q.pop();//清空佇列
    while(tree[i].f!=-1) i++;//找根節點的位子
    q.push(tree[i]);
    for(int depth=1;1;depth++)
    {
        int t=q.size();//t表示哈夫曼樹中深度為depth的節點的數量
        if(t==0) return ;//t等於0時，哈夫曼樹每一層所有節點都遍歷過了
        while(t--)
        {
            struct node now=q.front();
            q.pop();
            if(now.ls==-1&&now.rs==-1)
            {
                *p+=(depth-1)*now.v;
            }
            else
            {
                if(now.ls!=-1)
                    q.push(tree[now.ls]);
                if(now.rs!=-1)
                    q.push(tree[now.rs]);


            }
        }
    }
}
int main()
{
    int ans=0;//帶權路徑長度
    int index1,index2;
    cin>>num;
    memset(tree,-1,sizeof(tree));
    for(int i=0;i<num;i++)
        cin>>tree[i].v;
    while(judge())//judge函式用於判斷哈夫曼樹是否已經完成建立
    {
        int i=0;
        index1=-1;
        index2=-1;
        while(tree[i].f!=-1) i++;
            index1=i;
        for(i++;i<num;i++)
        {
            if(tree[i].f==-1)
            {
                if(tree[index1].v>tree[i].v)
                {
                    index2=index1;
                    index1=i;
                }
                else
                {
                    if(index2==-1)
                        index2=i;
                    else
                        if(tree[index2].v>tree[i].v)
                            index2=i;



                }
            }
        }
        tree[num].v=tree[index1].v+tree[index2].v;
        tree[num].ls=index1;
        tree[num].rs=index2;
        tree[index1].f=num;
        tree[index2].f=num;
        num++;

    }
    for(int i=0;i<num;i++)
        cout<<"index:"<<i<<"   node value:"<<tree[i].v<<"  ls:"<<tree[i].ls<<"  rs:"<<tree[i].rs<<"  f:"<<tree[i].f<<endl;
    bfs(&ans);//求解wpl
    cout<<"wpl:"<<ans<<endl;
    return 0;
}