// 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: 7.628931591999334x):
#include <bits/stdc++.h>

using namespace std;

#define INF 2000000000

typedef long long int ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

#define getbit(n,i) (((n)&(1<<(i)))!=0)

#define setbit0(n,i) ((n)&(~(1<<(i))))

#define setbit1(n,i) ((n)(1<<(i)))

#define togglebit(n,i) ((n)^(1<<(i)))

#define lastone(n) ((n)&(-(n)))

#define read freopen("debug\\in.txt","r",stdin)

#define write freopen("debug\\out.txt","w",stdout)

#define fi first

#define se second

#define getI(a) scanf("%d",&a)

#define getII(a,b) scanf("d",&a,&b)

#define PI (acos(-1))

#define fastread ios_base::sync_with_stdio(false);cin.tie(NULL)

#define mod (1000000007)

#define asz 500005

#define t int t;scanf("%d",&t);for(int v=1;v<=t;v++)

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

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

#define minheap priority_queue<int,vector<int>,greater<int> >

#define maaheap priority_queue<int>

#define pb push_back

#define newl "\n"

ll BIGMOD ( ll a, ll p, ll m )

{

    ll result = 1;

    ll x = a;



    while ( p )

    {

        if ( p & 1 ) //p is odd

        {

            result = ( result * x ) % m;

        }

        x = ( x * x ) % m;

        p = p >> 1;

    }



    return result;

}



bool isPrime(int n)

{

    if (n <= 1)  return false;

    if (n <= 3)  return true;

    if (n%2 == 0 || n%3 == 0) return false;



    for (int i=5; i*i<=n; i=i+6)

        if (n%i == 0 || n%(i+2) == 0)

           return false;



    return true;

}



void SieveOfEratosthenes(int n)

{

    bool prime[n+1];

    memset(prime, true, sizeof(prime));



    for (int p=2; p*p<=n; p++)

    {



        if (prime[p] == true)

        {

            for (int i=p*p; i<=n; i += p)

                prime[i] = false;

        }

    }

}

const int MAX=100005;

ll memo[MAX][3];

 int arr[MAX][3];

 int n;

 void init()

 {

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

     {

         for(int j=0;j<3;j++) memo[i][j]=-1;

     }

 }

ll vacation(int i,char current)

{

    if(i==n) return 0;

    if(memo[i][current-'A']!=-1) return memo[i][current-'A'];

    if(current=='A')

    {

        ll res1=arr[i][0]+vacation(i+1,'B');

        ll res2=arr[i][0]+vacation(i+1,'C');

        return memo[i][0]=max(res1,res2);

    }

    if(current=='B')

    {

        ll res1=arr[i][1]+vacation(i+1,'A');

        ll res2=arr[i][1]+vacation(i+1,'C');

        return memo[i][1]=max(res1,res2);

    }

    if(current=='C')

    {

        ll res1=arr[i][2]+vacation(i+1,'A');

        ll res2=arr[i][2]+vacation(i+1,'B');

        return memo[i][2]=max(res1,res2);

    }





}



int main()

{

  // fastread;

   cin>>n;

   init();

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

   {

       for(int j=0;j<3;j++) cin>>arr[i][j];

   }

   ll res1=vacation(0,'A');

   ll res2=vacation(0,'B');

   ll res3=vacation(0,'C');

   cout<<max(res1,max(res2,res3))<<newl;

  return 0;

}