Problem:
Score : 100 points

Problem Statement

Taro's summer vacation starts tomorrow, and he has decided to make plans for it now.

The vacation consists of N days.
For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day:

A: Swim in the sea. Gain a_i points of happiness.

B: Catch bugs in the mountains. Gain b_i points of happiness.

C: Do homework at home. Gain c_i points of happiness.

As Taro gets bored easily, he cannot do the same activities for two or more consecutive days.

Find the maximum possible total points of happiness that Taro gains.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq a_i, b_i, c_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N

Output

Print the maximum possible total points of happiness that Taro gains.

Sample Input 1

3
10 40 70
20 50 80
30 60 90

Sample Output 1

210

If Taro does activities in the order C, B, C, he will gain 70 + 50 + 90 = 210 points of happiness.

Sample Input 2

1
100 10 1

Sample Output 2

100

Sample Input 3

7
6 7 8
8 8 3
2 5 2
7 8 6
4 6 8
2 3 4
7 5 1

Sample Output 3

46

Taro should do activities in the order C, A, B, A, C, B, A.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define ll              long long
#define pb              push_back
#define ppb             pop_back
#define endl            '\n'
#define mii             map<ll,ll>
#define pii             pair<ll,ll>
#define vi              vector<ll>
#define vs              vector<string>
#define all(a)          (a).begin(),(a).end()
#define F               first
#define S               second
#define sz(x)           (ll)x.size()
#define hell            1000000007
#define inf             (ll)1e18 + 5
#define PI              3.14159265358979323844
#define rep(i,a,b)      for (ll i=a;i<b;i++)
#define repr(i,a,b)     for (ll i=a-1;i>=b;i--)
#define lbnd            lower_bound
#define ubnd            upper_bound
#define mp              make_pair
#define ios             ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ordered_set     tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define trace(...)      __f(#__VA_ARGS__, __VA_ARGS__)

template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');std::cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}

using namespace std;
using namespace __gnu_pbds;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//primes for hashing 937,991,1013,1409,1741

ll n;
vi a,b,c;
ll dp[100005][3];

ll find(ll n,ll i){
	if (n<=0)return 0;
	if (dp[n][i]!=-1)return dp[n][i];
	if (i==0){
		dp[n][i]=a[n-1];
		ll meow=max(find(n-1,1),find(n-1,2));
		dp[n][i]+=max(meow,find(n-2,0));
	}
	else if (i==1){
		dp[n][i]=b[n-1];
		ll meow=max(find(n-1,0),find(n-1,2));
		dp[n][i]+=max(meow,find(n-2,1));
	}
	else {
		dp[n][i]=c[n-1];
		ll meow=max(find(n-1,1),find(n-1,0));
		dp[n][i]+=max(meow,find(n-2,2));
	}
	return dp[n][i];
}

void solve() {
    cin>>n;
    memset(dp,-1,sizeof dp);
    rep(i,0,n){
    	ll x,y,z;
    	cin>>x>>y>>z;
    	a.pb(x),b.pb(y),c.pb(z);
    }
    find(n,0);
    find(n,1);
    find(n,2);
    cout<<max(dp[n][0],max(dp[n][1],dp[n][2]))<<endl;
}

signed main() {
    ios
    ll test=1;
    //cin>>test;
    while (test--) solve();
    return 0;
}