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: Wrong Answer)

#include <bits/stdc++.h>

#define ld long double
#define ll long long int
#define ull unsigned long long int
#define vi vector<int>
#define vl vector<ll>
#define vvi vector< vector<int> >
#define vvl vector< vector<ll> >
#define repd(i, a, b) for (int i=(a);i<(b);i++)
#define rep(i, n) repd(i,0,n)
#define ALL(v) v.begin(), v.end()
#define INF 1e9

using namespace std;

/**
 * calculate GCD(greatest common divisor)
 * @param a
 * @param b
 * @return
 */
unsigned euclidean_gcd(unsigned a, unsigned b) {
    if (a < b) return euclidean_gcd(b, a);
    unsigned r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

/**
 * change minimum
 * @tparam T
 * @param a
 * @param b
 * @return bool
 */
template<class T>
inline bool changeMinimum(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

/**
 * change maximum
 * @tparam T
 * @param a
 * @param b
 * @return bool
 */
template<class T>
inline bool changeMaximum(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

/**
 * output answer
 * @tparam T
 * @param answer
 * @return
 */
template<typename T>
T output(T answer) {
    cout << answer << endl;
    return 0;
}

/** ----from here ---------------------------------------------------------- */

int main() {

    int N;
    cin >> N;

    // 入力
    long long a[100010][3]; // a[i], b[i], c[i] をそれぞれまとめて a[i][0], a[i][1], a[i][2] にしてしまう

    // DP テーブル
    long long dp[100010][3];
    for (int i = 0; i < N; ++i) for (int j = 0; j < 3; ++j) cin >> a[i][j];


    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < 3; ++j) {
            for (int k = 0; k < 3; ++k) {
                if (j == k) { continue; }
//                changeMaximum(dp[i + 1][k], dp[i][j] + a[i][k]);
                                changeMaximum(dp[i + 1][k], dp[i][j] + a[i][k]);


            }

        }
    }
    // 答え
    long long res = 0;
    for (int j = 0; j < 3; ++j) changeMaximum(res, dp[N][j]);
    cout << res << endl;


}