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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

//const int mod = 1e9+9;

/*ll gcd(ll x, ll y){return (y ? gcd(y, x%y) : x); }

*ll lcm(ll x, ll y){return x*(y/gcd(x,y));}

*void exEuclid(int a, int b){

	if(b==0){x = 1; y=0; d=a; return ;}

	exEuclid(b, a%b);

	int x1=y;

	int y1=x-(a/b)*y;

	x=x1; 100000

	y=y1;

} 

* */

//===========================================================//

int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i){

	if(i == n) return 0;

	if(dp[a][i] != -1) return dp[a][i];

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>n;

	for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

	cout<<max(go(0,0),max(go(1,0),go(2,0)))<<"\n";

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.204683417714469x):
#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;

}