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>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define pf push_front
#define eb emplace_back
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define SZ(x) (int)(x).size()
#define ft front()
#define bk back()
#define endl '\n'
#define FOR(a, b, c) for (auto(a) = (b); (a) < (c); ++(a))
#define ROF(a, b, c) for (auto(a) = (b); (a) > (c); --(a))
#define F0R(a, b) FOR ((a), 0, (b))
#define R0F(a, b) ROF ((a), (b), -1)
#define CEIL(a, b) ((a) + (b)-1) / (b)
#define SET(a, b) memset((a), (b), sizeof(a))
template <class t>
bool ckmin(t& a, const t& b) {
    return a > b ? a = b, true : false;
}
template <class t>
bool ckmax(t& a, const t& b) {
    return a < b ? a = b, true : false;
}
#ifndef DEBUG
#define DEBUG 0
#endif
#define dout \
    if (DEBUG) std::cerr
#define dvar(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
template <class a, class b>
ostream& operator<<(ostream& o, const pair<a, b>& p);
template <class T>
struct IsC {
    template <class C>
    static char test(typename C::const_iterator*);
    template <class C>
    static int test(...);
    static const bool value = sizeof(test<T>(0)) == sizeof(char);
};
template <>
struct IsC<string> {
    static const bool value = false;
};
template <class C>
typename enable_if<IsC<C>::value, ostream&>::type operator<<(ostream& o,
                                                             const C& c) {
    o << '{';
    for (auto it = c.begin(); it != c.end();)
        o << *it << (++it != c.end() ? ", " : "}");
    return o;
}
template <class a, class b>
ostream& operator<<(ostream& o, const pair<a, b>& p) {
    return o << '(' << p.fi << ", " << p.se << ')';
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int n;
    cin >> n;
    vvi vs(n, vi(3));
    F0R (i, n)
        cin >> vs[i][0] >> vs[i][1] >> vs[i][2];
    vvi dp(n, vi(3, 0));
    for (int i = 0; i < 3; ++i) dp[0][i] = vs[0][i];

    for (int i = 1; i < n; ++i)
        for (int j = 0; j < 3; ++j)
            for (int k = 0; k < 3; ++k)
                if (k != j) ckmax(dp[i][j], dp[i - 1][k] + vs[i][j]);

    dout << dvar(dp) << endl;

    cout << *max_element(ALL(dp[n - 1])) << endl;

    return 0;
}