// 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: 2.6165268484320188x):
//Bismillahir Rahmanir Rahim

 

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

//#pragma GCC target("avx,avx2,fma")

        

#include<bits/stdc++.h>

#include<ext/pb_ds/assoc_container.hpp>

#include<ext/pb_ds/tree_policy.hpp>

 

#define ll long long

#define ull unsigned long long

#define pii pair<int,int>

#define pll pair<ll,ll>

#define xx first

#define yy second

#define sci(n) scanf("%d",&n)

#define scii(n,m) scanf("%d%d",&n,&m)

#define sciii(n,m,w) scanf("%d%d%d",&n,&m,&w)

#define scl(n) scanf("%lld",&n)

#define scll(n,m) scanf("%lld%lld",&n,&m)

#define sclll(n,m,w) scanf("%lld%lld%lld",&n,&m,&w)

#define pf(a)   printf("%d\n",a)

#define CASE(a) printf("Case %d: ",a)

#define endl '\n'

#define READ  freopen("input.txt","r",stdin)

#define WRITE freopen("output.txt","w",stdout);

#define PI acos(-1)

#define mem(a,b) memset(a,b,sizeof(a))

#define SQR(a) (a)*(a)

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

#define pb push_back

#define pri priority_queue<int>

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

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

#define mod 1000000007

#define inf INT_MAX

#define eps 1e-9

 

using namespace __gnu_pbds;

using namespace std;



#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }



void err(istream_iterator<string> it) {}

template<typename T, typename... Args>

void err(istream_iterator<string> it, T a, Args... args) {

  cerr << *it << " = " << a << endl;

  err(++it, args...);

}

 

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

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

 

//ll ncr(ll n,ll r){if(n==r)return 1;if(r==1)return n;if(dp[n][r]!=-1)return dp[n][r];return dp[n][r]=ncr(n-1,r)+ncr(n-1,r-1);}

 

// bit manipulations

 

//bool checkbit(int mask,int bit){return mask & (1<<bit);}

//int setbit(int mask,int bit){ return mask | (1<<bit) ; }

//int clearbit(int mask,int bit){return mask & ~(1<<bit);}

//int togglebit(int mask,int bit){return mask ^ (1<<bit);}

//int bitno(int mask) {return (int)__builtin_popcount(mask);}

 

template <typename T> T BigMod (T b,T p,T m){if (p == 0) return 1;if (p%2 == 0){T s = BigMod(b,p/2,m);return ((s%m)*(s%m))%m;}return ((b%m)*(BigMod(b,p-1,m)%m))%m;}

template <typename T> T ModInv (T b,T m){return BigMod(b,m-2,m);}

template <typename T> T POW(T B,T P){ if(P==0) return 1; if(P&1) return B*POW(B,P-1);  else return SQR(POW(B,P/2));}

template <typename T> T Dis(T x1,T y1,T x2, T y2){return sqrt( SQR(x1-x2) + SQR(y1-y2) );}

template <typename T> T Angle(T x1,T y1,T x2, T y2){ return atan( double(y1-y2) / double(x1-x2));}

template <typename T> T gcd(T a,T b){if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}

template <typename T> T lcm(T a,T b) {if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}

template <typename T> T euclide(T a,T b,T &x,T &y) {if(a<0){T d=euclide(-a,b,x,y);x=-x;return d;}   if(b<0){T d=euclide(a,-b,x,y);y=-y;return d;}   if(b==0){x=1;y=0;return a;}else{T d=euclide(b,a%b,x,y);T t=x;x=y;y=t-(a/b)*y;return d;}}

template <typename T> void ia (T a[],int n){for (int i=0; i<n; i++) cin >> a[i];}

template <typename T> void pa (T a[],int n){for (int i=0; i<n-1; i++) cout << a[i] << " ";cout << a[n-1] << endl;}

template <typename T> ll isLeft(T a,T b,T c) { return (a.x-b.x)*(b.y-c.y)-(b.x-c.x)*(a.y-b.y);}

 

template<typename T>

using ordered_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

 

///********************************MY CODE STARTS HERE********************************///

 

 

#define mxn 100005



int a[mxn] , b[mxn] , c[mxn];

int dp[3][mxn];

 

void solve(){

    

    int n;

    cin >> n;



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

    	cin >> a[i] >> b[i] >> c[i];

    }



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

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

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

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



    }



    cout << max({dp[0][n] , dp[1][n] , dp[2][n]}) << endl;







}

 

 

int main(){

       //IOS;

    

    int T = 1;

    // cin >> T;

    while(T--){

        solve();

    }

 

return 0;

}