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



using namespace std;



const int N = 1e5 + 9, OO = 0x3f3f3f3f;



int n, points[N][3], dp[N][4];



int solve(int i, int prev){

    if(i == n)  return 0;



    int &ret = dp[i][prev];

    if(~ret)    return ret;



    ret = 0;

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

        if(prev != j)   ret = max(ret, points[i][j] + solve(i + 1, j));





    return ret;

}



int main(){



    cin >> n;

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

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



    memset(dp, -1, sizeof dp);



    cout << solve(0, 3);

}


// Faster C++ solution for the same problem (measured speedup: 2.602015945319543x):
//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;

}