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

#define ll long long

using namespace std;



ll n;

std::vector<ll> ara[100010];

ll dp[100010][4];



ll fnc(ll idx,ll last){

  if(idx == n) return 0;

  if(dp[idx][last] != -1) return dp[idx][last];



  ll res1 = - (1<<25);

  ll res2 = - (1<<25);

  ll res3 = - (1<<25);



  if(last == 0){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 1){

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 2){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 3){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

  }

  dp[idx][last] = max(res1, max(res2,res3));

  return dp[idx][last];

}



int main(int argc, char const *argv[]) {

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

  cin >> n;

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

    int a,b,c;

    cin >> a >> b >> c;

    ara[i].push_back(a);

    ara[i].push_back(b);

    ara[i].push_back(c);

  }



  cout << fnc(0,0) << endl;

  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 6.949484104719401x):
///#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;

    }

}