// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 2.62713380890139x):
#include <iostream>

#include <map>

#include <vector>

#include <algorithm>

#include <set>

#include <stdio.h>

#include <math.h>

#include <queue>

#include <stack>

using namespace std;

typedef long long int lli;

const long long int MaxN=200005;

const lli mod=1e9+7;

#define fst first

#define snd second

#define mp make_pair

const float pi=3.1415926535897932384626433832795028;

	lli mcd(lli a,lli b){

		if(a==0)return b;

		return mcd(b%a,a);

	}

	lli dp[MaxN][3];										//dp[día][A]=max(dp[día-1][C]+A[soy],dp[día-1][C]+A[soy];

	lli A[MaxN],B[MaxN],C[MaxN];

	void solve(){

		lli N,K,i,j,p,q;

		cin>>N;

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

			cin>>A[i]>>B[i]>>C[i];

		}

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

			dp[i][1]=max(dp[i-1][2],dp[i-1][3])+A[i];

			dp[i][2]=max(dp[i-1][1],dp[i-1][3])+B[i];

			dp[i][3]=max(dp[i-1][1],dp[i-1][2])+C[i];

		}

		lli res=dp[N][1];

		res=max(res,max(dp[N][2],dp[N][3]));

		cout<<res<<"\n";

	}

int main(){

	//cin.tie(0),ios_base::sync_with_stdio(0);

	lli i,j,t;

	t=1;

	for(i=1;i<=t;i++){

		solve();

	}

	

}