// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



typedef long double LD;

typedef long long ll;

typedef map<int,int> mii;

typedef pair<int,int> pii;

typedef vector<pii> vpii;



#define pb              push_back

#define mp              make_pair

#define F               first

#define S               second

#define mod             1000000007

#define INF             2147483647

#define Endl            "\n"

#define FasterIO        ios_base::sync_with_stdio(0);cin.tie(0); cout.tie(0) ;

#define precision(a)    cout << fixed << setprecision(a)

#define bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define reset(a,b)      memset(a,b,sizeof(a));

#define srt(a)      sort(a.begin(),a.end());



const int MAXN = 1000005;

const int SQRTN = 1003;

const int LOGN = 22;

const double PI=acos(-1);

const int sz=1e5+9;



int strtoint(string str){

    stringstream ss(str);

    int x = 0;

    ss >> x ;

    return x ;

}

string intostr(int x){

    stringstream ss;

    ss << x;

    string str = ss.str();

    return str;

}

void rem(string &s, const string &c) {

	auto pos = s.find(c);

	if (pos == string::npos) {

		cout << -1 << endl;

		exit(0);

	}

	s.erase(0, pos + 1);

}

void println (int n){

    cout<<n<<"\n";

}

void printvec(vector <int> &a){

    for (int i=0;i<a.size();i++){

        cout<<a[i]<<" ";

    }

    cout<<"\n";

}



void printgraph(vector <int> g[],int e){

    for (int i=0;i<e;i++){

        cout<<i<<" --> ";

        for (int j=0;j<g[i].size();j++){

            cout<<g[i][j]<<" ";

        }

        cout<<"\n";

    }

}



void printmap (map <int,int> &mymap){

    for (auto itr = mymap.begin(); itr != mymap.end(); ++itr) {

        cout << '\t' << itr->first

             << '\t' << itr->second << '\n';

    }

    cout << "\n";

}



int a[100005][4], dp[100005][4];

int N;



int rec(int pos, int type){



    if(pos>N) return 0;



    if(dp[pos][type]!=-1) return dp[pos][type];



    int ret = -1;

    for(int i=1; i<=3; i++)

        if(i!=type)

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



    return dp[pos][type] = ret;

}





void solve(){

    memset(dp,-1,sizeof(dp));

    int i,j;

    cin>>N;

    for(i=1; i<=N; i++)

        for(j=1; j<=3; j++)

            cin>>a[i][j];





   cout<< rec(0,0) << "\n";



    //printvec(v);

}





int main(){

    //FasterIO;



    solve();

    return 0;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 9;

inline int read()

{

    int data = 0, f = 1;

    char ch = getchar();

    while (!isdigit(ch))

    {

        if (ch == '-')

            f = -1;

        ch = getchar();

    }

    while (isdigit(ch))

    {

        data = (data << 3) + (data << 1) + ch - '0';

        ch = getchar();

    }

    return f * data;

}

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main()

{

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

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

    //std::ios::sync_with_stdio(false);

    //std::cin.tie(0);

    int n = read();

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

    {

        a[i] = read(), b[i] = read(), c[i] = read();

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

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

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

    }

    printf("%d\n",max(max(dp[n][0], dp[n][1]), dp[n][2]));

    //fclose(stdin);

    //fclose(stdout);

    return 0;

}


// Fastest known solution for this problem: