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

using namespace std;

typedef long long ll;



ll arr[100009][3];

int dp[100009][2][2][2];



long long VAC1(int n,int a, int b, int c)

{



    if(!n && a) return arr[n][0];

    else if(!n && b) return arr[n][1];

    else if(!n && c) return arr[n][2];

    else if(dp[n][a][b][c]!=-1)return dp[n][a][b][c];



    else

    {



        ll x,y,z;

        if(a)

        {

            x = VAC1(n-1,0,1,0)+arr[n][0];

            y = VAC1(n-1,0,0,1)+arr[n][0];



        }



        else if(b)

        {

            x = VAC1(n-1,1,0,0)+arr[n][1];

            y = VAC1(n-1,0,0,1)+arr[n][1];

        }



        else if(c)

        {

            x = VAC1(n-1,1,0,0)+arr[n][2];

            y = VAC1(n-1,0,1,0)+arr[n][2];

        }



        dp[n][a][b][c] = max(x,y);;

        return dp[n][a][b][c];





    }

}



int main()

{

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

    int n;

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

    cin >> n;

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

    {

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

        {

            cin >> arr[i][j];

        }

    }

    ll res = VAC1(n-1,1,0,0);

    res= max(res, VAC1(n-1,0,1,0));

    res = max(res,VAC1(n-1,0,0,1));



    cout << res << endl;

}


// Faster C++ solution for the same problem (measured speedup: 4.093935857345811x):
///#pragma GCC optimize("Ofast")

///#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")

///#pragma GCC optimize("unroll-loops")



#include<bits/stdc++.h>

using namespace std;



///#include<ext/pb_ds/assoc_container.hpp>

///#include<ext/pb_ds/tree_policy.hpp>

///using namespace __gnu_pbds;

///template<typename T> using orderset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

///(X).order_of_key(value) /// return lower_bound(value)

///(*X).find_by_order(index) /// return value (0 index)



mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); /// mt19937_64 (long long)



typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

typedef vector<int> vii;

typedef vector<ll> vll;

typedef vector<vll> vvll;

typedef vector<vii> vvii;

typedef pair<int,int> pii;

typedef pair<ll,ll> pll;

typedef vector<pii> vpii;

typedef vector<pll> vpll;

typedef vector<bool> vbb;



const int N = 1000001;

const int MOD = 1000000007;

const ll INF = LLONG_MAX;



#define fast() ios_base::sync_with_stdio(false),cin.tie(NULL)

#define next(x) next_permutation(all(x))

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

#define rall(a) a.rbegin(),a.rend()

#define rev(a) reverse(all(a))

#define eb emplace_back

#define pb push_back

#define mp make_pair

#define ff first

#define ss second

#define yes cout<<"YES"<<nl

#define no cout<<"NO"<<nl

#define space ' '

#define nl '\n'

#define PI acos(-1)

#define Case(x) cout<<"Case "<<x<<": "

#define strtoint(a) atoi(a.c_str())

#define dbg(x) cerr<<#x<<" is "<<x<<'\n';

#define fix(x) cout<<fixed<<setprecision(x)

#define leastonepos(x) __builtin_ffs(x)

#define leadingoff(x) __builtin_clz(x)

#define trailingoff(x) __builtin_ctz(x)

#define numofone(x) __builtin_popcount(x)

#define time() cerr<<"Time elapsed : "<<clock()*1000.0/CLOCKS_PER_SEC<<"ms"<<'\n'



#define coutv(v) for(auto it:v)cout<<it<<space;cout<<nl;

#define cinv(v) for(auto &it:v)cin>>it;



///................Graph's Move.................

///const int dx[] = {+1,-1,+0,+0}; ///Rock's Move

///const int dy[] = {+0,+0,+1,-1}; ///Rock's Move

///const int dx[] = {+0,+0,+1,-1,-1,+1,-1,+1}; ///King's Move

///const int dy[] = {-1,+1,+0,+0,+1,+1,-1,-1}; ///king's Move

///const int dx[] = {-2,-2,-1,-1,+1,+1,+2,+2}; ///knight's Move

///const int dy[] = {-1,+1,-2,+2,-2,+2,-1,+1}; ///knight's Move

///*.....................-_-........................*///







int main()

{

    fast();

    int tc=1,cs=0;

    //cin>>tc;

    while(tc--)

    {

        int n,i,k,j;

        cin>>n;

        vvii v(n,vii(3)),dp(n,vii(3));

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

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

                cin>>v[i][j];

        dp[0][0]=v[0][0];

        dp[0][1]=v[0][1];

        dp[0][2]=v[0][2];

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

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

                for(k=0; k<3; k++)

                {

                    if(j==k)

                        continue;

                    dp[i][k]=max(dp[i][k],dp[i-1][j]+v[i][k]);

                }

        int ans=*max_element(all(dp[n-1]));

        cout<<ans<<nl;

    }

}