// ------- PBDS ------ //
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
// ------- PBDS ------ //

#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
#define SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define ll long long
#define testcase ll tt; cin>>tt; while(tt--)
#define pb push_back
#define nn cout<<"\n"
#define MOD 1000000007
#define pi 3.1415926535
#define endl "\n"
#define vll vector<ll>
#define MULTI int T; cin >> T; while(T--)
#define all(a)      (a).begin(),(a).end()
#define here cout<<"I'm here\n";
void vout(vector<ll> ans){
    for(auto ele:ans) cout<<ele<<" "; cout<<endl;
}
vector<ll> adj[200001];
bool vis[200001];

// ------- PBDS ------ //
typedef tree<
// change this for any data-type
string,
null_type,
// change this for any data-type
less<string>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;

/*
    to get the position of an element in set , do
    s.order_of_key(ele);
    this returns the number of elements strictly lesser than ele in set s
    Let us assume we have a set s : {1, 5, 6, 17, 88}, then :
    s.order_of_key(6) : Count of elements strictly smaller than 6 is 2.
    s.order_of_key(25) : Count of elements strictly smaller than 25 is 4.

    to get the element at position k
    s.find_by_order(k);
    returns an iterator to the ele counting from 0
    Let us assume we have a set s : {1, 5, 6, 17, 88}, then :
    *(s.find_by_order(2)) : 3rd element in the set i.e. 6
    *(s.find_by_order(4)) : 5th element in the set i.e. 88
*/

// ------- PBDS ------ //

clock_t time_p=clock();
void purvansh11()
{
    time_p=clock()-time_p;
    cerr<<"Time Taken : "<<(float)(time_p)/CLOCKS_PER_SEC<<"\n";
}

//BINARY EXPONENTIATION  -- O(log2(b))
ll power(ll a, ll b) {
    /*
        for 3^11 :
        as 11 -> 1011, we mean 3^8 * 3^2 * 3^1
        so multiply for all ones in binary rep of b
    */

    ll ans = 1;
    ll num = a;
    while (b > 0) {
        if (b & 1) ans = ((ans % MOD) * (num % MOD)) % MOD;
        num = ((num % MOD) * (num % MOD)) % MOD;
        b /= 2;
    }
    return ans;
}
ll n;
vector<ll> a,b,c;
ll dp[100005][3][3][3];
ll solve(ll i,ll same_a,ll same_b,ll same_c){
    if(i==n) return 0;
    if(dp[i][same_a][same_b][same_c]!=-1) return dp[i][same_a][same_b][same_c]; 
    ll x,y,z;
    x=-1e9;
    y=-1e9;
    z=-1e9;
    if(same_a < 1)
        x = a[i] + solve(i+1 , same_a + 1 , 0 , 0);
    if(same_b < 1)
        y = b[i] + solve(i+1 , 0 , same_b + 1 , 0);
    if(same_c < 1)
        z = c[i] + solve(i+1 , 0 , 0 , same_c + 1);

    return dp[i][same_a][same_b][same_c] = max(x,max(y,z)); 
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    SPEED;
    cout.precision(8);
    cout << fixed;
    cin>>n;
    memset(dp,-1,sizeof(dp));
    for(ll i=0;i<n;i++){
        ll ta,tb,tc;
        cin>>ta>>tb>>tc;
        a.pb(ta);
        b.pb(tb);
        c.pb(tc);
    }
    cout<<solve(0,0,0,0);
    purvansh11();
    return 0;
}

Measured memory_bytes: 39304