#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MAXN = 10010;
const int MAXM = 1001000;
const int INF = 1<<30;
struct EDG{
    int to,next,cap,flow;
    int cost;  //单位价格
}edg[MAXM];
int head[MAXN],eid;
int pre[MAXN], cost[MAXN] ; //点0~(n-1)

void init(){
    eid=0;
    memset(head,-1,sizeof(head));
}
void addEdg(int u,int v,int cap,int cst){
    edg[eid].to=v; edg[eid].next=head[u]; edg[eid].cost = cst;
    edg[eid].cap=cap; edg[eid].flow=0; head[u]=eid++;

    edg[eid].to=u; edg[eid].next=head[v]; edg[eid].cost = -cst;
    edg[eid].cap=0; edg[eid].flow=0; head[v]=eid++;
}

bool inq[MAXN];
bool spfa(int sNode,int eNode , int n){
    queue<int>q;
    for(int i=0; i<n; i++){
        inq[i]=false; cost[i]= -1;
    }
    cost[sNode]=0; inq[sNode]=1; pre[sNode]=-1;
    q.push(sNode);
    while(!q.empty()){
        int u=q.front(); q.pop();
        inq[u]=0;
        for(int i=head[u]; i!=-1; i=edg[i].next){
            int v=edg[i].to;
            if(edg[i].cap-edg[i].flow>0 && cost[v]<cost[u]+edg[i].cost){ //在满足可增流的情况下，最小花费
                cost[v] = cost[u]+edg[i].cost;
                pre[v]=i;   //记录路径上的边
                if(!inq[v])
                    q.push(v),inq[v]=1;
            }
        }
    }
    return cost[eNode]!=-1;    //推断有没有增广路
}
//反回的是最大流，最小花费为minCost
int minCost_maxFlow(int sNode,int eNode ,int& minCost , int n){
    int ans=0;
    while(spfa(sNode,eNode , n)){

        for(int i=pre[eNode]; i!=-1; i=pre[edg[i^1].to]){
            edg[i].flow+=1; edg[i^1].flow-=1;
            minCost+=edg[i].cost;
        }
        ans++;
        if(ans==2)break;
    }
    return ans;
}
int main(){
    int n,mapt[35][35];
    while(scanf("%d",&n)>0){
        init();
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            scanf("%d",&mapt[i][j]);
        int s = 0 , t = n*n-1;

        for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
        if(i||j){
            addEdg(i*n+j , i*n+j+n*n , 1 , mapt[i][j]);
            if(j+1<n) addEdg(i*n+j+n*n, i*n+j+1 , 1 , 0 );
            if(i+1<n) addEdg(i*n+j+n*n, (i+1)*n+j , 1 , 0);
        }
        else{
            addEdg(s , 1 , 1,0) , addEdg(s , n , 1 , 0);
        }
        int maxCost=mapt[0][0];
        if(n>1) maxCost+=mapt[n-1][n-1];
        minCost_maxFlow(s , t , maxCost , n*n*2);
        printf("%d\n",maxCost);
    }
}