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

using namespace std;

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

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

typedef long long ll;

typedef unsigned long long ull;

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

#define forr(i, a, b) for(ll i = (a); i < (ll) (b); i++)

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define MOD 1000000007

#define MAXN 100005

ll dp[MAXN][3];

void solve(vector< vector<ll> > &a, int n)

{

    forn(i,MAXN)

    {

        forn(j,3)

        {

            dp[i][j]=(pow(10,9)+5)*(ll)(-1);

        }

    }

    dp[0][0]=a[0][0], dp[0][1]=a[0][1], dp[0][2]=a[0][2];

    forr(i,1,n)

    {

        forn(j,3)

        {

            forn(k,3)

            {

                if(j==k)

                {

                    continue;

                }

                else dp[i][j]=max(dp[i][j],dp[i-1][k]+a[i][j]);

            }

        }

 

    }

    cout << max(dp[n-1][0],max(dp[n-1][1],dp[n-1][2])) << "\n";

}

 

int main() {

    FIN;

    int n;

    cin >> n;

    vector< vector<ll> > arreglo(n);

    forn(i,n) arreglo[i].resize(3);

    forn(i,n)

    {

        forn(j,3) cin >> arreglo[i][j];

    }

    solve(arreglo, n);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.37725665613304x):
#pragma region

#include <algorithm>

#include <cstring>

#include <iostream>

#include <map>

#include <queue>

#include <stack>

#include <string>

#include <vector>

typedef long long ll;

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define rep(i, a, n) for (long long i = a; i <= n; ++i)

#define per(i, a, n) for (long long i = n; i >= a; --i)

#define IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)

template <typename T>

void read(T &x) {

    x = 0;

    char ch = getchar();

    ll f = 1;

    while (!isdigit(ch)) {

        if (ch == '-') f *= -1;

        ch = getchar();

    }

    while (isdigit(ch)) {

        x = x * 10 + ch - 48;

        ch = getchar();

    }

    x *= f;

}

using namespace std;

/*

 *                                                     __----~~~~~~~~~~~------___

 *                                    .  .   ~~//====......          __--~ ~~

 *                    -.            \_|//     |||\\  ~~~~~~::::... /~

 *                 ___-==_       _-~o~  \/    |||  \\            _/~~-

 *         __---~~~.==~||\=_    -_--~/_-~|-   |\\   \\        _/~

 *     _-~~     .=~    |  \\-_    '-~7  /-   /  ||    \      /

 *   .~       .~       |   \\ -_    /  /-   /   ||      \   /

 *  /  ____  /         |     \\ ~-_/  /|- _/   .||       \ /

 *  |~~    ~~|--~~~~--_ \     ~==-/   | \~--===~~        .\

 *           '         ~-|      /|    |-~\~~       __--~~

 *                       |-~~-_/ |    |   ~\_   _-~            /\

 *                            /  \     \__   \/~                \__

 *                        _--~ _/ | .-~~____--~-/                  ~~==.

 *                       ((->/~   '.|||' -_|    ~~-/ ,              . _||

 *                                  -_     ~\      ~~---l__i__i__i--~~_/

 *                                  _-~-__   ~)  \--______________--~~

 *                                //.-~~~-~_--~- |-------~~~~~~~~

 *                                       //.-~~~--\

 *                       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 */



#pragma endregion

const int maxn = 1e5 + 5;

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main() {

    IO;

    int n;

    cin >> n;

    rep(i, 1, n) cin >> a[i] >> b[i] >> c[i];

    rep(i, 1, n) {

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

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

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

    }

    cout << max({dp[n][0], dp[n][1], dp[n][2]}) << endl;

}