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)

#include <bits/stdc++.h>
using namespace std;
 
//long long
using ll = long long;
 
// pair<int, int>
using PII = pair<int, int>;
 
 
//最大値、mod
const int MOD = 1000000007;
const int mod = 1000000007;
const int INF = 1000000000;
const long long LINF = 1e18;
const int MAX = 510000;
 
//出力系
#define print(x) cout << x << endl
#define prints(x) cout << fixed << setprecision(20) << x << endl
#define printc(x) cout << setw(2) << setfill('0') << x << endl;
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
 
//配列入力
vector<long long>vecin(ll n){
    vector<long long>res(n);
    for(int i = 0; i < n; i++) cin >> res[i];
    return res;
}

// begin() end()
#define all(x) (x).begin(),(x).end()
 
//for
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define rep(i,a,b) for(int i=(a);i<(b);i++)
 
//最大公約数 
 ll gcd(ll x, ll y) { return y ? gcd(y,x%y) : x;}

// 最小公倍数
unsigned lcm(unsigned a, unsigned b){
    return a / gcd(a, b) * b; 
}
 
// a = max(a, b), a = min(a, b)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
// 階乗(MODをとる)
ll pow_mod(ll num, ll pow, ll mod) {
    ll prod = 1;
    num %= mod;
    while (pow > 0) {
        if (pow & 1) prod = prod * num % mod;
        num = num * num % mod;
        pow >>= 1;
    }
    return prod;
}
 
 
 
// 二項係数(MODとる、1 ≦ k ≦ n ≦ 10^7 程度)
// COMinit()
// COM(x, y)
// とコンビで使う
 
// テーブルを作る前処理
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 

 
 
// UnionFind
struct UnionFind {
    vector<int> par;
    vector<int> rank;
    vector<ll> Size;
    UnionFind(int n = 1) {
        init(n);
    }
 
    void init(int n = 1) {
        par.resize(n + 1); rank.resize(n + 1); Size.resize(n + 1);
        for (int i = 0; i <= n; ++i) par[i] = i, rank[i] = 0, Size[i] = 1;
    }
 
    int root(int x) {
        if (par[x] == x) {
            return x;
        }
        else {
            int r = root(par[x]);
            return par[x] = r;
        }
    }
 
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
 
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        Size[x] += Size[y];
        return true;
    }
 
    ll size(int x){
        return Size[root(x)];
    }
};

 
//modint構造体 
struct Mint {
    int val;
    Mint inv() const{
        int tmp,a=val,b=mod,x=1,y=0;
        while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
        return Mint(x);
    }
public:
    Mint():val(0){}
    Mint(ll x){if((val=x%mod)<0)val+=mod;}
    Mint pow(ll t){Mint res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;}return res;}
    Mint& operator+=(const Mint& x){if((val+=x.val)>=mod)val-=mod;return *this;}
    Mint& operator-=(const Mint& x){if((val+=mod-x.val)>=mod)val-=mod; return *this;}
    Mint& operator*=(const Mint& x){val=(ll)val*x.val%mod; return *this;}
    Mint& operator/=(const Mint& x){return *this*=x.inv();}
    bool operator==(const Mint& x) const{return val==x.val;}
    bool operator!=(const Mint& x) const{return val!=x.val;}
    bool operator<(const Mint& x) const{return val<x.val;}
    bool operator<=(const Mint& x) const{return val<=x.val;}
    bool operator>(const Mint& x) const{return val>x.val;}
    bool operator>=(const Mint& x) const{return val>=x.val;}
    Mint operator+(const Mint& x) const{return Mint(*this)+=x;}
    Mint operator-(const Mint& x) const{return Mint(*this)-=x;}
    Mint operator*(const Mint& x) const{return Mint(*this)*=x;}
    Mint operator/(const Mint& x) const{return Mint(*this)/=x;}
};
struct factorial {
    vector<Mint> Fact, Finv;
public:
   //factorial fact(10000010);
   //fact.nCr(a, b)
  //「fact」の部分は自由に名前変更可能
    factorial(int maxx){
        Fact.resize(maxx+1),Finv.resize(maxx+1); Fact[0]=Mint(1); rep(i,0,maxx)Fact[i+1]=Fact[i]*(i+1);
        Finv[maxx]=Mint(1)/Fact[maxx]; rrep(i,maxx,0)Finv[i-1]=Finv[i]*i;
    }
    Mint fact(int n,bool inv=0){if(inv)return Finv[n];else return Fact[n];}
    Mint nPr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[n-r];}
    Mint nCr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[r]*Finv[n-r];}
};


// 1 * 2 * 3 .... * n (mod)
ll modfact(ll n) { 
    if (n <= 1) return 1;
    return (n * modfact(n - 1)) % MOD;
}


// kが角度だった場合:cos(k * (PI / 180));
//const double PI = acos(-1);のまま使うと円周率(M_PIもあるよ)
const double PI = acos(-1);


// 多次元 vector 生成   例: auto dp = make_vec<long long>(N+1, 5, 5, 5); 
template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}
template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
  return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}

//素因数分解
vector<pair<long long, int>>factorize(long long n){
  vector<pair<long long, int>> res;
  for(long long i = 2; i * i <= n; ++i){
    if(n % i) continue;
    res.emplace_back(i, 0);
    while(n % i == 0){
      n /= i;
      res.back().second++;
    }
  }
if(n != 1) res.emplace_back(n, 1);
return res;
}

// 素数判定
bool primejudge(long long a){
    if(a <= 1) return false;
    for(long long i = 2; i * i <= a; i++){
        if(a % i == 0) return false;
    }
    return true;
}

// // 累積和
// vector<long long>sums(vector<int>n){
//     vector<long long>res(n.size() + 1, 0);
//     for(int i = 0; i < n.size(); i++) res[i + 1] = n[i] + res[i];
//     return res;
// }

int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};



int main(){
    int n;
    cin >> n;
    vector<vector<int>>dp(n + 1, vector<int> (3, 0));
    REP(i, n){
        int a, b, c;
        cin >> a >> b >> c;
        if(i == 0) {
            dp[i + 1][0] = dp[i][0] + a;
            dp[i + 1][1] = dp[i][1] + b;
            dp[i + 1][2] = dp[i][2] + c;
        }else{
            dp[i + 1][0] = a + max(dp[i][1], dp[i][2]);
            dp[i + 1][1] = b + max(dp[i][0], dp[i][2]);
            dp[i + 1][2] = c + max(dp[i][1], dp[i][0]);
        }
    }
    cout << max({dp[n][0], dp[n][1], dp[n][2]}) << endl;
    return 0;
}