// 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: 2.9914695584724087x):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#include <complex>

#define ll long long

#define  point complex<double>

#define  pi acos(-1)

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

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const ll N = 2e5 + 5, mod = 1e9 + 7, inff = -1e12,inf=1e12;

const double ep = 0.0000000001;

//const double pi = 3.14159265359;



double dot(point a, point b) {

	return  (conj(a)*b).real();

}

double cross(point a, point b) {

	return  (conj(a)*b).imag();

}

double angle(point a, point b)

{

	double dt = dot(a, b);

	dt /= abs(a);

	dt /= abs(b);

	return acos(dt) * 180.0 / pi;

}



bool isrightangel(int ind) {//90 degree

	vector<point>v;

	return(angle(v[ind] - v[(ind + 1) % 3], v[ind] - v[(ind + 2) % 3]) == 90);//bshof hal eldot producty3ny fe

	//dl3en 3moden 3ashn yb2a right trinagle

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

	if (exp == 0)

		return 1;

	ll ans = fp(base, exp / 2);

	ans = (ans * ans) % mod;

	if (exp % 2 != 0)

		ans = (ans * (base % mod)) % mod;

	return ans;

}



void calcFacAndInv(ll n)

{

	fact[0] = inv[0] = 1;

	for (ll i = 1; i <= n; i++)

	{

		fact[i] = (i * fact[i - 1]) % mod;

		inv[i] = fp(fact[i], mod - 2);

	}

}



ll ncr(ll n, ll r)

{

	return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

	return (fact[n] * inv[n - r]) % mod;

}













	int dpa[100005], dpb[100005], dpc[100005];

int main()

{

	ll n,arr[100005],a,b,c;

	char tmp;

	cin >> n;

	

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

		cin >> dpa[i] >> dpb[i] >> dpc[i];

	}

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

		dpa[i] += max(dpb[i - 1], dpc[i - 1]);

		dpb[i] += max(dpa[i-1], dpc[i-1]);

		dpc[i] += max(dpb[i-1], dpa[i-1]);

	}

	cout << max(dpa[n - 1], max(dpb[n - 1], dpc[n - 1]));



	return 0;

}