// Slower C++ solution:
#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

int main()

{

    int n,i,j,k;

    int a[100005][5];

    scanf("%d",&n);

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

    scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

    int dp[100005][5];

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

    dp[1][1] = a[1][1];

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

    dp[1][3] = a[1][3];

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

    {

        for(j = 1;j < 4;j++)

        {

            for(k = 1;k < 4;k++)

            {

                if(k != j)

                {

                    if(dp[i][j] < dp[i - 1][k] + a[i][j])

                    dp[i][j] = dp[i - 1][k] + a[i][j];

                }

            }

        }

    }

    printf("%d\n",max(dp[n][1],max(dp[n][2],dp[n][3])));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2029879886018318x):
//                                 Alisher_2211

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        # include <bits/stdc++.h>

# define  long          long long

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

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

# define  pb            push_back

# define  fi            first 

# define  se            second

# define  pi            pair <long, long>    

# define  lcm(a, b)     (a * b) / __gcd(a, b)

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        using namespace std;

const long N = 100100;

void read(){freopen("input.txt" , "r" , stdin );freopen("output.txt", "w" , stdout );}

long dp[N][3];

long n, k;

void solve(){

    cin >> n;

    long ar[n + 5][5];

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

    {

        for (int j = 0; j < 3; ++j)

        {

            cin >> ar[i][j];

        }

    }

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

    {

        for (int j = 0; j < 3; ++j)

        {

            if( !j ) 

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

            if( j == 1 ) 

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

            else 

                dp[i][2] = max( dp[i - 1][1], dp[i - 1][0] ) + ar[i][2];// <=

        }

    }

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

}

int32_t main(){

ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); 

        long T = 1;

        //cin >> T;

        while( T-- ){

            solve();

            cout<<'\n';

        }

cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

}