// Slower C++ solution:
/*

██╗  ██╗ █████╗ ██████╗ ███████╗███████╗███╗   ███╗        ██╗  ██╗██╗  ██╗ █████╗ ██╗     ███████╗██████╗

██║ ██╔╝██╔══██╗██╔══██╗██╔════╝██╔════╝████╗ ████║        ██║ ██╔╝██║  ██║██╔══██╗██║     ██╔════╝██╔══██╗

█████╔╝ ███████║██████╔╝█████╗  █████╗  ██╔████╔██║        █████╔╝ ███████║███████║██║     █████╗  ██║  ██║

██╔═██╗ ██╔══██║██╔══██╗██╔══╝  ██╔══╝  ██║╚██╔╝██║        ██╔═██╗ ██╔══██║██╔══██║██║     ██╔══╝  ██║  ██║

██║  ██╗██║  ██║██║  ██║███████╗███████╗██║ ╚═╝ ██║███████╗██║  ██╗██║  ██║██║  ██║███████╗███████╗██████╔╝

╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═╝     ╚═╝╚══════╝╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═════╝

Believe in yourself ,( try and try and then try ), You can do it 

*/

#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<unordered_map>

#define endl "\n"

#define Ceil(x,y) ((x+y-1)/y)

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

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

#define allr(s) s.rbegin(),s.rend()

#define rt(s) return cout<<s,0

#define watch(x) cout<<(#x)<<" = "<<x<<endl

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

#define OO 0x3f3f3f3f3f3f3f3fLL

const int oo = 0x3f3f3f3f, mod = 1e9 + 7;

using namespace std;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

//x = xDegrees*3.14159265 / 180;

//ef1tn=(( (n / 2)*( (n / 2) + 1) / 2) * 2);

//of1tn=(n * (n + 1) / 2 - 2 * ((n / 2) * (n / 2 + 1) / 2));

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

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

void K_K(){

#ifndef ONLINE_JUDGE

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

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

#endif  !ONLINEJUDGE

	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

}

ll gcd(ll a, ll b) {

	return b == 0 ? abs(a) : gcd(b, a % b);

}

ll lcm(ll a, ll b) {

	return a / gcd(a, b) * b;

}

int n;

vector<int>a, b, c;

int dp[100005][5];

int sol(int idx, int pre){

	if (idx == n)

		return 0;

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

	if (~ret)

		return ret;

	int ch1 = 0, ch2 = 0, ch3 = 0;

	if (pre != 1)

		ch1 = sol(idx + 1, 1) + a[idx];

	if (pre != 2)

		ch2 = sol(idx + 1, 2) + b[idx];

	if (pre != 3)

		ch3 = sol(idx + 1, 3) + c[idx];

	return ret = max({ ch1, ch2, ch3 });

}

int main(){

	K_K();

	cin >> n;

	a.resize(n);

	b.resize(n);

	c.resize(n);

	clr(dp, -1);

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

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

	cout << sol(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 2.9560733458775883x):
#include<bits/stdc++.h>

using namespace std;



#define ff              first

#define ss              second

#define ll             long long

#define pb              push_back

#define mp              make_pair

#define pii             pair<ll,ll>

#define vi              vector<ll>

#define mii             map<ll,ll>

#define pqb             priority_queue<ll>

#define pqs             priority_queue<ll,vi,greater<ll> >

#define setbits(x)      __builtin_popcountll(x)

#define zrobits(x)      __builtin_ctzll(x)

#define mod             1000000007

#define inf             1e18

#define ps(x,y)         fixed<<setprecision(y)<<x

#define mk(arr,n,type)  type *arr=new type[n];

#define w(x)            ll x; cin>>x; while(x--)

#define FIO             ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)



struct activity{

ll A,B,C;};



ll solve(ll n,vector<activity>v){

    ll dp[n+1][3];

dp[1][0]=v[1].A;

dp[1][1]=v[1].B;

dp[1][2]=v[1].C;

for(int i=2;i<=n;i++)

{

    dp[i][0]=v[i].A+max(dp[i-1][1],dp[i-1][2]);

    dp[i][1]=v[i].B+max(dp[i-1][0],dp[i-1][2]);

    dp[i][2]=v[i].C+max(dp[i-1][1],dp[i-1][0]);

}



return max(dp[n][0],max(dp[n][1],dp[n][2]));



}



int main()

{

    FIO;

    ll a,b,c,d,e,f=0,sum=0,ans,ck=0,m,n,k,p,q,r,t,add=0,x,y,z;



    string s;



    cin>>n;

    vector<activity>v(n+1);

    for(int i=1;i<=n;i++){cin>>v[i].A;cin>>v[i].B;cin>>v[i].C;}

    ans=solve(n,v);

    cout<<ans<<endl;







    return 0;

}