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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
class Program
{
    //static List<string> str = new List<string>();
	static void Main(string[] args)
	{
		// 入力フェーズ
		int N = Int32.Parse(Console.ReadLine()); // 夏休みの日数
		//string[] stone = Console.ReadLine().Split(' '); // 足場の高さ
		int[] an = new int[N];
		int[] bn = new int[N];
		int[] cn = new int[N]; // a,b,cの幸福度
		for(int i=0; i<N; i++){
			string[] happy = Console.ReadLine().Split(' '); // 幸福度
			an[i] = Int32.Parse(happy[0]);
			bn[i] = Int32.Parse(happy[1]);
			cn[i] = Int32.Parse(happy[2]);
		}
//Console.WriteLine(height.Count + "");
		long[,] dp = new long[N+1,3]; //i日目に活動？をした時の幸福度の総和の最大値テーブル

// 初期値設定
		dp[0,0] = 0;
		dp[0,1] = 0;
		dp[0,2] = 0;
		for(int i=1; i<=N; i++)
		{
			// i日目に活動Aをした場合の幸福度の総和の最大値
			dp[i,0] = Math.Max(
										dp[i-1,1] + an[i-1] // 前日にBをした時の最大値＋今日Aをした時の幸福度
										, dp[i-1,2] + an[i-1] // 前日にCをした時の・・・
										);

			// i日目に活動Bをした時の幸福度の総和の最大値
			dp[i,1] = Math.Max(
										dp[i-1,0] + bn[i-1] // 前日にAをした時の最大値＋今日Bをした時の幸福度
										, dp[i-1,2] + bn[i-1] // 前日にCをした時の・・・
										);
			// i日目に活動Cをした時の幸福度の総和の最大値
			dp[i,2] = Math.Max(
										dp[i-1,1] + cn[i-1] // 前日にBをした時の最大値＋今日Cをした時の幸福度
										, dp[i-1,0] + cn[i-1] // 前日にAをした時の・・・
										);

			//Console.WriteLine(i + "日目; " + dp[i,0] + " " + dp[i, 1] + " " + dp[i,2]);
		}

		// dp[N][0,1,2]のうちの最大のものを出力する
		long maxHappy = Math.Max(dp[N,0], Math.Max(dp[N,1], dp[N,2]));
		Console.WriteLine(maxHappy + "");


    }

}