// Original C++ solution (slower):
/**

 *    Creator :  bizarre

 *    Date : 4/2020

 *    Problem Number : 424

**/

#include <bits/stdc++.h>

using namespace std;

#define bizarre     ios_base::sync_with_stdio( false ), cin.tie( nullptr ), cout.tie( nullptr );

/*

vector<int> prime_Numbers;

int primes[1000005];

void prime(int n)

{

    primes[1]=primes[0]=0;

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

        primes[i]=true;

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

    {

        if(primes[i]==true)

        {

            prime_Numbers.push_back(i);

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

                primes[j]=false;

        }

    }

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

        primes[i]+=primes[i-1];

}



void move1step(long long& a, long long& b, long long q)

{

    long long c= a-q*b;

    a=b;

    b=c;

}

long long egcd(long long r0,long long r1,long long& x0,long long& y0)

{

    long long x1=0,y1=1;

    x0=1;

    y0=0;

    while(r1)

    {

        long long q=r0/r1;

        move1step(r0,r1,q);

        move1step(x0,x1,q);

        move1step(y0,y1,q);

    }

    return r0;

}

bool  solve (long long a, long long b, long long c,long long&x, long long& y)

{

    long long g=egcd(a,b,x,y);

    int m=c/g;

    x*=m;

    y*=m;

    return c%g==0;

}







int npr(long long n, long long r)

{

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

}



int gcd(int a, int b)

{

    if(b==0)

        return a;

    return gcd(b,a%b);

}

void PrimeFactorization(int n)

{

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

    {

        int c=0;

        if(n%i==0)

        {

            while(n%i==0)

            {

                n/=i;

            }



        }

    }



}

const int inf = 2e9, mod = 1e9 + 7,N=1e5+5;

long long fact[N], inv[N];

long long fp(long long base, long long exp)

{

    if (exp == 0)

        return 1;

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

    ans = (ans * ans) % mod;

    if (exp % 2 != 0)

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

    return ans;

}

void pre(long long n)

{

    fact[0] = 1;

    inv[0] = 1;

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

    {

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

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

    }

}

int ncr(long long n, long long r)

{

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

}

long long ncrtane(long long n, long long r)

{

    long long ans=1,cntr=1;

    for(int i=n;i>n-r;i--)

    {

        ans*=i;

        ans/=cntr;

        cntr++;

    }

    return ans;

}

*/

int main()

{

    bizarre

     int n;

     cin>>n;

     long long a[n],b[n],c[n],aa[n],bb[n],cc[n];

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

     {

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

         if(i==0)

         {

             aa[i]=a[i];

             bb[i]=b[i];

             cc[i]=c[i];

         }

         else

         {

             aa[i]=max(cc[i-1],bb[i-1])+a[i];

             bb[i]=max(cc[i-1],aa[i-1])+b[i];

             cc[i]=max(aa[i-1],bb[i-1])+c[i];

         }

     }

     cout << max(aa[n-1],max(bb[n-1],cc[n-1]));

    return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

    for(i=0;i<N;i++)

    {

        scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]);

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

    printf("%d\n",max(d[N-1][0],max(d[N-1][1],d[N-1][2])));



    return 0;

}


// Fastest known solution for this problem: