//费用流模板题，因为题目中的表述就是一个网络流模型，拆点建图，没点只经过一次，起点终点容量为2，
//其他点容量为1，要求最大费用费用为负权值，套模板。起点和终点经历了两次，最后要减去。
/****************最小费用最大流模板,白书363页*******************************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int maxn=30*30*2+5,inf=0x7fffffff;//本体拆点，数组多开两倍
struct Edge
{
    int from,to,cap,flow,cost;
    Edge(int u,int v,int c,int f,int cs):from(u),to(v),cap(c),flow(f),cost(cs){}
};
struct MCMF
{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int>g[maxn];
    int inq[maxn],d[maxn],p[maxn],a[maxn];
    void init(int n)
    {
        this->n=n;
        for(int i=0;i<n;i++) g[i].clear();
        edges.clear();
    }
    void AddEdge(int from,int to,int cap,int cost)
    {
        edges.push_back((Edge){from,to,cap,0,cost});
        edges.push_back((Edge){to,from,0,0,-cost});
        m=edges.size();
        g[from].push_back(m-2);
        g[to].push_back(m-1);
    }
    bool BellmanFord(int s,int t,int &flow,int &cost)
    {
        for(int i=0;i<n;i++) d[i]=inf;
        memset(inq,0,sizeof(inq));
        d[s]=0;inq[s]=1;p[s]=0;a[s]=inf;
        queue<int>q;
        q.push(s);
        while(!q.empty()){
            int u=q.front();q.pop();
            inq[u]=0;
            for(int i=0;i<(int)g[u].size();i++){
                Edge &e=edges[g[u][i]];
                if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
                    d[e.to]=d[u]+e.cost;
                    p[e.to]=g[u][i];
                    a[e.to]=min(a[u],e.cap-e.flow);
                    if(!inq[e.to]) {q.push(e.to);inq[e.to]=1;}
                }
            }
        }
        if(d[t]==inf) return false;
        flow+=a[t];
        cost+=d[t]*a[t];
        int u=t;
        while(u!=s){
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }
    int Mincost(int s,int t)
    {
        int flow=0,cost=0;
        while(BellmanFord(s,t,flow,cost));
        return cost;//返回最小费用，flow存最大流
    }
}MC;
/**********************************************************************************/
int main()
{
    int n,mp[35][35];
    while(scanf("%d",&n)==1){
        int s=0,t=n*n*2+1;
        MC.init(n*n*2+2);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&mp[i][j]);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                int id=(i-1)*n+j;
                if(id==1){
                    MC.AddEdge(id,id+n*n,2,-mp[i][j]);
                    MC.AddEdge(s,id,2,0);
                }
                else if(id==n*n){
                    MC.AddEdge(id,id+n*n,2,-mp[i][j]);
                    MC.AddEdge(id+n*n,t,2,0);
                }
                else MC.AddEdge(id,id+n*n,1,-mp[i][j]);
                if(i<n){
                    int nid=id+n;
                    MC.AddEdge(id+n*n,nid,1,0);
                }
                if(j<n){
                    int nid=id+1;
                    MC.AddEdge(id+n*n,nid,1,0);
                }
            }
        }
        int ans=-(MC.Mincost(0,n*n*2+1)+mp[1][1]+mp[n][n]);
        printf("%d\n",ans);
    }
    return 0;
}

//和HDU2686一样，只是数据变大了，上一个模板会超内存,这个板不会。
/***********************最小费用最大流模板2*************************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=610*610*2+2;
const int maxm=4*maxn;//！边数要够
const int inf=0x7fffffff;
struct Edge
{
    int to,next,cap,flow,cost;
}edges[maxm];
int head[maxn],tol,pre[maxn],dis[maxn];
bool vis[maxn];
int N;
void init(int n)
{
    N=n;
    tol=0;
    memset(head,-1,sizeof(head));
}
void AddEdge(int u,int v,int cap,int cost)
{
    edges[tol].to=v;
    edges[tol].cap=cap;
    edges[tol].cost=cost;
    edges[tol].flow=0;
    edges[tol].next=head[u];
    head[u]=tol++;
    edges[tol].to=u;
    edges[tol].cap=0;
    edges[tol].cost=-cost;
    edges[tol].flow=0;
    edges[tol].next=head[v];
    head[v]=tol++;
}
bool spfa(int s,int t)
{
    queue<int>q;
    for(int i=0;i<=N;i++){
        dis[i]=inf;
        vis[i]=0;
        pre[i]=-1;
    }
    dis[s]=0;
    vis[s]=1;
    q.push(s);
    while(!q.empty()){
        int u=q.front();q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edges[i].next){
            int v=edges[i].to;
            if(edges[i].cap>edges[i].flow&&dis[v]>dis[u]+edges[i].cost){
                dis[v]=dis[u]+edges[i].cost;
                pre[v]=i;
                if(!vis[v]) {vis[v]=1;q.push(v);}
            }
        }
    }
    if(pre[t]==-1) return 0;
    return 1;
}
int MinCostFlow(int s,int t)
{
    int flow=0,cost=0;
    while(spfa(s,t)){
        int Min=inf;
        for(int i=pre[t];i!=-1;i=pre[edges[i^1].to])
            Min=min(Min,edges[i].cap-edges[i].flow);
        for(int i=pre[t];i!=-1;i=pre[edges[i^1].to]){
            edges[i].flow+=Min;
            edges[i^1].flow-=Min;
            cost+=edges[i].cost*Min;
        }
        flow+=Min;
    }
    return cost;//返回最小费用，flow存最大流
}
/*********************************************************************/
int main()
{
    int n,mp[605][605];
    while(scanf("%d",&n)==1){
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++) scanf("%d",&mp[i][j]);
        int s=0,t=n*n*2+1;
        init(n*n*2+2);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                int id=(i-1)*n+j;
                if(id==1){
                    AddEdge(id,id+n*n,2,-mp[i][j]);
                    AddEdge(s,id,2,0);
                }
                else if(id==n*n){
                    AddEdge(id,id+n*n,2,-mp[i][j]);
                    AddEdge(id+n*n,t,2,0);
                }
                else AddEdge(id,id+n*n,1,-mp[i][j]);
                if(i<n) AddEdge(id+n*n,id+n,1,0);
                if(j<n) AddEdge(id+n*n,id+1,1,0);
            }
        }
        int ans=-(MinCostFlow(s,t)+mp[1][1]+mp[n][n]);
        printf("%d\n",ans);
    }
    return 0;
}