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++ (GCC 9.2.1) (judge verdict: Accepted)

//g++ t.cpp -o t && t < p.txt
//d>p.txt&&t<p.txt&&t2<p.txt

#include <iostream> // プログラムの耳と口
#include <vector> // 超有能配列秘書
#include <algorithm> // みんな大好きソートと二分探索
#include <queue> // きゅーちゃん、だいすき
#include <string> // to_string
#include <functional> // 関数を変数に入れる子
#include <set> // 値の取得・挿入・削除を高速に
#include <map> // setの妹 これまた優秀
#include <random> // 乱択さん
#include <cassert>

#define DB cerr<<"D"<<endl
using namespace std; using ll=long long; using ld=long double; const int INF=1e9; const ll LINF=1e18; const double dINF = 1e18; const ld ldINF = 1e18; const double EPS = 1e-6;
template<typename T, typename U, typename O> void caut(T a, U b, O c){cout<<"("<<a<<","<<b<<","<<c<<") ";} template<typename T, typename U> void caut(T a, U b){cout<<"("<<a<<","<<b<<") ";} template<typename T> void caut(T a){cout<<"("<<a<<") ";};
using P=pair<ll,ll>;
const ll M = 1e9+7;
ll mod_pow(ll x, ll a) { ll an = 1; while(a > 0) { if (a&1) an = an * x % M; x = x * x % M; a >>= 1;} return an;}
void add(ll& x, ll y) {x+=y; x%=M;}; void mul(ll& x, ll y) {x*=y; x%=M;}; template<typename T, typename U> void chmax(T& x, U y) {if (x<y) x=y;}; template<typename T, typename U> void chmin(T& x, U y) {if (x>y) x=y;}
bool vaild(int x, int y, int hh, int ww){return 0<=x&&x<hh&&0<=y&&y<ww;}
int keta(ll a) {int res=0; while(a>0) res+=(a%10/5)+(a%5), a/=10; return res;}

int main() {
  int n; cin>>n;
  ll a[n],b[n],c[n];
  for(int i=0;i<n;i++) {
    cin>>a[i]>>b[i]>>c[i];
  }
  ll dp[n][3];fill(dp[0],dp[n],0);
  dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

  for(int i=1;i<n;i++) {
    dp[i][0] = max(dp[i-1][1], dp[i-1][2]) + a[i];
    dp[i][1] = max(dp[i-1][0], dp[i-1][2]) + b[i];
    dp[i][2] = max(dp[i-1][0], dp[i-1][1]) + c[i];
  }
  cout<<max({dp[n-1][0],dp[n-1][1],dp[n-1][2]})<<endl;
}
// ・配列の大きさok？　・priority_queueはgreater？　・debug消した？
// ・落ち着いて。提出まで4分待ってね……WJ……1/10……2/10……

// Thank you for making problems and running the contest