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

typedef long long ll;

using namespace std;

const int N = 2e5+10 , M = 5e5+10, OO = 0x3f3f3f3f ,  MOD = 1e9+7;

const double EPS = 1e-3;

#define M_PI 3.14159265358979323846

#define clr(arr,val) memset(arr , val , sizeof (arr))

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

#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)

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

#define pb push_back

#define F first

#define S second

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

#define sz(v)  ((int)((v).size()))

typedef pair <int,int> pii ;

typedef vector < pii > vii ;

typedef vector < int > vi;

int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction

//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction

//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction

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

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







int n,x;

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

ll dp[N][4];





ll solve(int idx, int preC){ //preColour = (1 : a) -- (2 : b) -- (3 : c) -- (0 : invalid)



    if(idx == n) return 0;



    ll& ret = dp[idx][preC];

    if(~ret) return ret;



    ret = -OO;

    if(preC != 1) ret = max(ret, solve(idx+1 , 1)+a[idx]);

    if(preC != 2) ret = max(ret, solve(idx+1 , 2)+b[idx]);

    if(preC != 3) ret = max(ret, solve(idx+1 , 3)+c[idx]);



    return ret;



}







int main(){





    //ios_base::sync_with_stdio(0);

    //cin.tie(0);cout.tie(0);





    scanf("%d",&n);

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

        scanf("%d %d %d",a+i,b+i,c+i);



    memset(dp, -1 , sizeof dp);

    ll ans = solve(0 , 0);

    printf("%lld\n",ans);







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.738734172145297x):
#include<bits/stdc++.h>

using namespace std;

 

#define             ll  long long

#define            ull  unsigned long long

#define            P_B  push_back

#define            M_P  make_pair

#define         DBG(x)  printf("%s %d\n", #x, x)

#define       DBGLL(x)  printf("%s %lld\n", #x, x)

#define         SCI(x)  scanf("%d",&x)

#define         SCD(x)  scanf("%lf",&x)

#define         SCS(x)  scanf("%s",x)

#define         CLR(x)  memset(x, 0, sizeof(x))

#define        CLRN(x)  memset(x, -1, sizeof(x))

#define        SCLL(x)  scanf("%lld", &x)

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

#define  FORR(i, m, n)  for(int i = m; i < n; i++)

#define     SCI2(m, n)  scanf("%d%d", &m, &n)

#define  SCI3(m, n, o)  scanf("%d%d%d",&m, &n, &o)

#define     SCD2(m, n)  scanf("%lf%lf",&m,&n)

#define  SCD3(m, n, o)  scanf("%lf%lf%lf",&m,&n,&o)

#define    SCLL2(m, n)  scanf("%lld%lld", &m, &n)

#define SCLL3(m, n, o)  scanf("%lld%lld%lld",&m, &n, &o)

#define             vi  vector<int>

#define            vii  vector<pair<int, int> >

#define            vll  vector<pair<ll, ll> >

#define            Mod  998244353

#define            INF  1000000000

 



int a[100005][4];

int n;



int main()

{

	SCI(n);

	FOR(i, n)

		FOR(j, 3)

			SCI(a[i][j]);

	FORR(i, 1, n){

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

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

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

	}

	printf("%d\n", max(a[n-1][0], max(a[n-1][1] , a[n-1][2])));    

}