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

#include <algorithm>

#include <iostream>

#include <iomanip>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <vector>

#include <string>

#include <climits>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <deque>

#include <stdio.h>

//#define int long long

using namespace std;

#define loop(n) for(int i=0;i<n;i++)

#define endl '\n'

#define OnlineJudgec

#define F first

#define S second

#define  precision(n)  cout << fixed << setprecision(n)

#define clr(v,d ) memset(v, d, sizeof(v))

typedef long long ll;

typedef vector<int> vi;

void init()

{

    cin.tie(0);

    cin.sync_with_stdio(0);

    cout.tie(0);

    #ifdef OnlineJudge

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

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

    #endif

}

ll S(ll x){ll s = 0;while(x){s += x % 10;x /= 10;}return s;} /// sum of digits.

ll powe(ll a , ll n){if(n == 1)return a;return a * powe(a ,n-1);}

ll com(int n,int m){ll z = 1;for(int i = 0; i < m;i++){z = z * ( n - i) / (i + 1) ;}return z;}

int dx[] = {1,-1,0,0};

int dy[] = {0,0,1,-1};

int dr[] = {0,0,0,1,1,1,-1,-1,-1};

int dc[] ={-1,0,1,-1,0,1,-1,0,1};

/// <<------------------------------------------------------------------------------>>

const int mx = 100009;

int n , k;

int arr[mx][4];

int dp[mx][4];

int solve(int idx , int last)

{

     if(idx == n)return 0;

     int &ret = dp[idx][last];

     if(~ret) return ret;

     int A = 0 ,B = 0 ,C = 0;

     if(last != 0)

        A = solve(idx+1 , 0) + arr[idx][0];

     if(last != 1)

        B = solve(idx+1 , 1) + arr[idx][1];

     if(last != 2)

        C = solve(idx+1 , 2) + arr[idx][2];

     return ret = max(A,max(B,C));

}

/// <<------------------------------------------------------------------------------>>

int main()

{

    init();

    cin >> n;

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

    {

        cin >> arr[i][0] >> arr[i][1] >> arr[i][2];

    }

    clr(dp,-1);

    cout<<solve(0,3);

}


// Faster C++ solution for the same problem (measured speedup: 2.5871560937491953x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

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

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



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

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

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



return 0;

}