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

#include <algorithm>

#include <iostream>

#include <iomanip>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <vector>

#include <string>

#include <climits>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <deque>

#include <stdio.h>

//#define int long long

using namespace std;

#define loop(n) for(int i=0;i<n;i++)

#define endl '\n'

#define OnlineJudgec

#define F first

#define S second

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

#define clr(v,d ) memset(v, d, sizeof(v))

typedef long long ll;

typedef vector<int> vi;

void init()

{

    cin.tie(0);

    cin.sync_with_stdio(0);

    cout.tie(0);

    #ifdef OnlineJudge

    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);

    #endif

}

ll S(ll x){ll s = 0;while(x){s += x % 10;x /= 10;}return s;} /// sum of digits.

ll powe(ll a , ll n){if(n == 1)return a;return a * powe(a ,n-1);}

ll com(int n,int m){ll z = 1;for(int i = 0; i < m;i++){z = z * ( n - i) / (i + 1) ;}return z;}

int dx[] = {1,-1,0,0};

int dy[] = {0,0,1,-1};

int dr[] = {0,0,0,1,1,1,-1,-1,-1};

int dc[] ={-1,0,1,-1,0,1,-1,0,1};

/// <<------------------------------------------------------------------------------>>

const int mx = 100009;

int n , k;

int arr[mx][4];

int dp[mx][4];

int solve(int idx , int last)

{

     if(idx == n)return 0;

     int &ret = dp[idx][last];

     if(~ret) return ret;

     int A = 0 ,B = 0 ,C = 0;

     if(last != 0)

        A = solve(idx+1 , 0) + arr[idx][0];

     if(last != 1)

        B = solve(idx+1 , 1) + arr[idx][1];

     if(last != 2)

        C = solve(idx+1 , 2) + arr[idx][2];

     return ret = max(A,max(B,C));

}

/// <<------------------------------------------------------------------------------>>

int main()

{

    init();

    cin >> n;

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

    {

        cin >> arr[i][0] >> arr[i][1] >> arr[i][2];

    }

    clr(dp,-1);

    cout<<solve(0,3);

}


// Faster C++ solution for the same problem (measured speedup: 2.613333786441925x):
#include <bits/stdc++.h>

using namespace std;

#define INF 2000000000

typedef long long int ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

#define getbit(n,i) (((n)&(1<<(i)))!=0)

#define setbit0(n,i) ((n)&(~(1<<(i))))

#define setbit1(n,i) ((n)(1<<(i)))

#define togglebit(n,i) ((n)^(1<<(i)))

#define lastone(n) ((n)&(-(n)))

#define read freopen("debug\\in.txt","r",stdin)

#define write freopen("debug\\out.txt","w",stdout)

#define fi first

#define se second

#define getI(a) scanf("%d",&a)

#define getII(a,b) scanf("d",&a,&b)

#define PI (acos(-1))

#define fastread ios_base::sync_with_stdio(false);cin.tie(NULL)

#define mod (1000000007)

#define asz 500005

#define t int t;scanf("%d",&t);for(int v=1;v<=t;v++)

#define loop_arr(i,a,b) for(int i=a;i<b;i++)

#define loop_norm(i,a,b) for(int i=a;i<=b;i++)

#define minheap priority_queue<int,vector<int>,greater<int> >

#define maaheap priority_queue<int>

#define pb push_back

#define newl "\n"

ll BIGMOD ( ll a, ll p, ll m )

{

    ll result = 1;

    ll x = a;



    while ( p )

    {

        if ( p & 1 ) //p is odd

        {

            result = ( result * x ) % m;

        }

        x = ( x * x ) % m;

        p = p >> 1;

    }



    return result;

}



bool isPrime(int n)

{

    if (n <= 1)  return false;

    if (n <= 3)  return true;

    if (n%2 == 0 || n%3 == 0) return false;



    for (int i=5; i*i<=n; i=i+6)

        if (n%i == 0 || n%(i+2) == 0)

           return false;



    return true;

}



void SieveOfEratosthenes(int n)

{

    bool prime[n+1];

    memset(prime, true, sizeof(prime));



    for (int p=2; p*p<=n; p++)

    {



        if (prime[p] == true)

        {

            for (int i=p*p; i<=n; i += p)

                prime[i] = false;

        }

    }

}

const int MAX=100005;

ll memo[MAX][3];

 int arr[MAX][3];

 int n;

 void init()

 {

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

     {

         for(int j=0;j<3;j++) memo[i][j]=-1;

     }

 }

ll vacation(int i,char current)

{

    if(i==n) return 0;

    if(memo[i][current-'A']!=-1) return memo[i][current-'A'];

    if(current=='A')

    {

        ll res1=arr[i][0]+vacation(i+1,'B');

        ll res2=arr[i][0]+vacation(i+1,'C');

        return memo[i][0]=max(res1,res2);

    }

    if(current=='B')

    {

        ll res1=arr[i][1]+vacation(i+1,'A');

        ll res2=arr[i][1]+vacation(i+1,'C');

        return memo[i][1]=max(res1,res2);

    }

    if(current=='C')

    {

        ll res1=arr[i][2]+vacation(i+1,'A');

        ll res2=arr[i][2]+vacation(i+1,'B');

        return memo[i][2]=max(res1,res2);

    }





}



int main()

{

  // fastread;

   cin>>n;

   init();

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

   {

       for(int j=0;j<3;j++) cin>>arr[i][j];

   }

   ll res1=vacation(0,'A');

   ll res2=vacation(0,'B');

   ll res3=vacation(0,'C');

   cout<<max(res1,max(res2,res3))<<newl;

  return 0;

}