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



using namespace std;



typedef long long ll;

const double eps=1e-6,pi=acos(-1);

const int N=1e5+5,M=1e9+7,SEGM=4*N,OO=0x3f3f3f3f;

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

int mem[N][5];



int solve(int i=0,int bk=0){

    if(i==n) return 0;

    int &ret=mem[i][bk];

    if(~ret) return ret;

    if(bk!=1)

        ret=max(ret,solve(i+1,1)+a[i]);

    if(bk!=2)

        ret=max(ret,solve(i+1,2)+b[i]);

    if(bk!=3)

        ret=max(ret,solve(i+1,3)+c[i]);

    return ret;

}



int main(){

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

    scanf("%d",&n);

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

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

    memset(mem,-1,sizeof mem);

    printf("%d",solve());

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.8987774835912408x):
//                                 Alisher_2211

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        # include <bits/stdc++.h>

# define  long          long long

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

# define  allr(vc)      vc.rbegin(),vc.rend()

# define  pb            push_back

# define  fi            first 

# define  se            second

# define  pi            pair <long, long>    

# define  lcm(a, b)     (a * b) / __gcd(a, b)

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        using namespace std;

const long N = 100100;

void read(){freopen("input.txt" , "r" , stdin );freopen("output.txt", "w" , stdout );}

long dp[N][3];

long n, k;

void solve(){

    cin >> n;

    long ar[n + 5][5];

    for (int i = 1; i <= n; ++i)

    {

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

        {

            cin >> ar[i][j];

        }

    }

    for (int i = 1; i <= n; ++i)

    {

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

        {

            if( !j ) 

                dp[i][0] = max( dp[i - 1][1], dp[i - 1][2] ) + ar[i][0];

            if( j == 1 ) 

                dp[i][1] = max( dp[i - 1][0], dp[i - 1][2] ) + ar[i][1];

            else 

                dp[i][2] = max( dp[i - 1][1], dp[i - 1][0] ) + ar[i][2];// <=

        }

    }

    cout << max({ dp[n][0], dp[n][1], dp[n][2] });

}

int32_t main(){

ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); 

        long T = 1;

        //cin >> T;

        while( T-- ){

            solve();

            cout<<'\n';

        }

cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

}