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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;
import java.util.function.Predicate;
 
public class Main {
 
    public static void main(String[] args) {
        new Main().run();
    }
 
    void run() {
        Scanner cin = new Scanner(System.in);
        N = cin.nextInt();
        A = new int[N];
        B = new int[N];
        C = new int[N];
        for (int i = 0; i < N; i ++){
            A[i] = cin.nextInt();
            B[i] = cin.nextInt();
            C[i] = cin.nextInt();
        }
        dp = new int[N][3];
        for (int i=0;i<N;i++){
            for (int j=0;j<3;j++){
                dp[i][j] = -1;
            }
        }
 
        System.out.println(dfs(0, -1));
    }
 
    int N;  //最大日数
    int[] A, B, C;  //それぞれの行動で得られる幸福度
    int[][] dp; // day, preChoiceの2引数に対する答えを記録する -1で初期化済み
    //何日目で最後に選んだのがどれの時に、これ以降最大どれだけの幸福度が得られるかを返す
    int dfs(int day, int preChoice) {
        //もし最終日であれば、もう幸福度は得られないので、0を返す
        if (day == N) return 0;
        //答えが計算済みであれば、dp配列から答えを返す
        if (dp[day][preChoice] != -1) return -1;
        //最終日でなければ、3つの選択肢から選ぶ
        int chooseA, chooseB, chooseC;
        //選べない場合の最低値を入れておく。
        chooseA = chooseB = chooseC = Integer.MIN_VALUE;
        //A, B, Cをそれぞれ選んだ場合の幸福度を計算する
        if (preChoice != 0) chooseA = A[day] + dfs(day + 1, 0);
        if (preChoice != 1) chooseB = B[day] + dfs(day + 1, 1);
        if (preChoice != 2) chooseC = C[day] + dfs(day + 1, 2);
        //最も幸福度が高かったものを返す dp配列に記録する
        return dp[day][preChoice] = Math.max(chooseA, Math.max(chooseB, chooseC));
    }
}