// Slower C++ solution:
#include <bits/stdc++.h>

typedef long long ll;

using namespace std;

const int N = 2e5+10 , M = 5e5+10, OO = 0x3f3f3f3f ,  MOD = 1e9+7;

const double EPS = 1e-3;

#define M_PI 3.14159265358979323846

#define clr(arr,val) memset(arr , val , sizeof (arr))

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

#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)

#define range(i,a,b) for(int i = int(a);i <= int(b);i++)

#define pb push_back

#define F first

#define S second

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

#define sz(v)  ((int)((v).size()))

typedef pair <int,int> pii ;

typedef vector < pii > vii ;

typedef vector < int > vi;

int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction

//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction

//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction

//freopen("input.txt", "r", stdin);

//freopen("output.txt", "w", stdout);







int n,x;

int a[N],b[N],c[N];

ll dp[N][4];





ll solve(int idx, int preC){ //preColour = (1 : a) -- (2 : b) -- (3 : c) -- (0 : invalid)



    if(idx == n) return 0;



    ll& ret = dp[idx][preC];

    if(~ret) return ret;



    ret = -OO;

    if(preC != 1) ret = max(ret, solve(idx+1 , 1)+a[idx]);

    if(preC != 2) ret = max(ret, solve(idx+1 , 2)+b[idx]);

    if(preC != 3) ret = max(ret, solve(idx+1 , 3)+c[idx]);



    return ret;



}







int main(){





    //ios_base::sync_with_stdio(0);

    //cin.tie(0);cout.tie(0);





    scanf("%d",&n);

    for(int i = 0 ; i < n ; ++i)

        scanf("%d %d %d",a+i,b+i,c+i);



    memset(dp, -1 , sizeof dp);

    ll ans = solve(0 , 0);

    printf("%lld\n",ans);







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.632134910299155x):


#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main()

{

        ll N;

        cin>>N;

        ll A[N]={},B[N]={},C[N]={};

        ll i,j,k;



      ll  bal[N][3]={};





        for(i=0;i<N;i++){

                        cin>>A[i]>>B[i]>>C[i];



                for(j=0;j<3;j++){

                                if(i==0){

                                        if(j==0)bal[i][0]=bal[i][0]+A[i];

                                        else if(j==1)bal[i][1]=bal[i][1]+B[i];

                                        else bal[i][2]=bal[i][2]+C[i];

                                }

                                else {

                                        for(k=0;k<3;k++){

                                                if(j==k)continue;

                                                else if(j==0){

                                                            bal[i][j]=max(bal[i][j],bal[i-1][k]+A[i]);

                                                                }

                                                else if(j==1){

                                                          bal[i][j]=max(bal[i][j],bal[i-1][k]+B[i]);

                                                }

                                                else{

                                                        bal[i][j]=max(bal[i][j],bal[i-1][k]+C[i]);

                                                }

                                        }

                                }

                }

        }

        //cout<<bal[N-1][0]<<" "<<bal[N-1][1]<<" "<<bal[N-1][2];



        ll xx=max(bal[N-1][0],bal[N-1][1]);

        xx=max(xx,bal[N-1][2]);



        cout<<xx<<endl;

        return 0;

}