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

using namespace std;

const int oo = 1e9;

int number;

const int nx = 1e5+5;

int ac[nx][3];

int dyn[3][nx];

int init(int a, int i){

	if(i == number) return 0;

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

	int& res = dyn[a][i];

	

	if(a==0){

		res=max(res, init(1, i+1)+ac[i][0]);

		res=max(res, init(2, i+1)+ac[i][0]);

	}

	if(a==1){

		res=max(res, init(0, i+1)+ac[i][1]);

		res=max(res, init(2, i+1)+ac[i][1]);

	}

	if(a==2){

		res=max(res, init(0, i+1)+ac[i][2]);

		res=max(res, init(1, i+1)+ac[i][2]);

	}

	

	return res;

}

int main(){

	cin>>number;

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

	memset(dyn, -1, sizeof dyn);

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

	return 0;

}




// Faster C++ solution for the same problem (measured speedup: 2.1435429650330264x):
//                                 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";

}