// 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: 2.9760715072565573x):
/* BISMILLAHIR RAHMANIR RAHIM */

#include<bits/stdc++.h>

using namespace std;

#define ll        long long

#define ull       unsigned long long

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

#define l0(i,n)   for(ll i=0; i<n; i++)

#define MAX       100005

#define maxn 	  200005

#define pb 		  push_back

#define mk 		  make_pair

#define endl	  "\n"

#define INF 	  1000000000

#define MOD		  1000000007

#define imie(...) "[" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "





int main()

{

	int n;

	cin >> n;

	vector<int> hap[n+5];

	vector<int> dp[n+5];

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

	{

		int a,b,c;

		cin >> a >> b >> c;

		hap[i].pb(a);

		hap[i].pb(b);

		hap[i].pb(c);

		dp[i].pb(0);

		dp[i].pb(0);

		dp[i].pb(0);

	}

	//memset(dp,0,sizeof(dp));

	dp[0][0] = hap[0][0];

	dp[0][1] = hap[0][1];

	dp[0][2] = hap[0][2];

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

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

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

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

	}

	int ans = *max_element(dp[n-1].begin(),dp[n-1].end());

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

//	{

//		ans = max(ans,dp[n-1][i]);

//	}

	cout << ans << endl;

	return 0;

}