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# (.NET Core 3.1.201) (judge verdict: Runtime Error)

using System;

public static class Solution
{
  private static int[,] s_arr;
  private static int[,] s_memo;

  public static void Main()
  {
    int n = int.Parse(Console.ReadLine());
    s_arr = new int[n, 3];
    s_memo = new int[n, 3];
    for (int i = 0; i < n; i++)
    {
      var ithDay = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
      s_arr[i, 0] = ithDay[0];
      s_arr[i, 1] = ithDay[1];
      s_arr[i, 2] = ithDay[2];
      s_memo[i, 0] = -1;
      s_memo[i, 1] = -1;
      s_memo[i, 2] = -1;
    }
    if (n == 1)
    {
      Console.WriteLine(Math.Max(s_arr[0, 0], Math.Max(s_arr[0, 1], s_arr[0, 2])));
    }
    else
    {
      int max1 = s_arr[0, 0] + Solve(0, 1); // Solve(lastTakenActivity, startingDay);
      int max2 = s_arr[0, 1] + Solve(1, 1); // Solve(lastTakenActivity, startingDay);
      int max3 = s_arr[0, 2] + Solve(2, 1); // Solve(lastTakenActivity, startingDay);
      Console.WriteLine(Math.Max(max1, Math.Max(max2, max3)));
    }
  }

  private static int Solve(int lastTakenActivity, int startingDay)
  {
    if (s_memo[startingDay, lastTakenActivity] != -1)
      return s_memo[startingDay, lastTakenActivity];
    
    int option1, option2, option1Index, option2Index;
    if (lastTakenActivity == 0)
    {
      option1 = s_arr[startingDay, 1];
      option1Index = 1;
      option2 = s_arr[startingDay, 2];
      option2Index = 2;
    }
    else if (lastTakenActivity == 1)
    {
      option1 = s_arr[startingDay, 0];
      option1Index = 0;
      option2 = s_arr[startingDay, 2];
      option2Index = 2;
    }
    else
    {
      option1 = s_arr[startingDay, 0];
      option1Index = 0;
      option2 = s_arr[startingDay, 1];
      option2Index = 1;
    }

    if (startingDay == s_arr.Length / 3 - 1) return Math.Max(option1, option2); // Array length returns n * m here. We want n only, so divide by m = 3.

    return s_memo[lastTakenActivity, startingDay] =
      Math.Max(
        option1 + Solve(option1Index, startingDay + 1),
        option2 + Solve(option2Index, startingDay + 1));
  }
}