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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.lang.*;
import java.io.*;

// SHIVAM GUPTA :
//NSIT
//decoder_1671
//BEING PERFECTIONIST IS NOT AN OPTION



// ASCII = 48 + i ;

// SHIVAM GUPTA :


/* Name of the class has to be "Main" only if the class is public. */



public class Main
{
     
     
     
    static PrintWriter out;
    
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			out=new PrintWriter(System.out);
		}
		String next(){
			while(st==null || !st.hasMoreElements()){
				try{
					st= new StringTokenizer(br.readLine());
				}
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		long nextLong(){
			return Long.parseLong(next());
		}
		double nextDouble(){
			return Double.parseDouble(next());
		}
		String nextLine(){
			String str = "";
			try{
				str=br.readLine();
			}
			catch(IOException e){
				e.printStackTrace();
			}
			return str;
		}
	}
	





/////////////////////////////////////////////////////////////////////////////////////

public static int  min(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[0];
       
}
 
 /////////////////////////////////////////////////////////////////////////////
 
 
public static int  max(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[3];
       
}
 
/////////////////////////////////////////////////////////////////////////////////// 
static int sieve =  1000000 ;

 
static boolean[] prime =  new boolean[sieve + 1] ;

 
public static void sieveOfEratosthenes() 
    { 
        // FALSE == prime
        
        // TRUE ==  COMPOSITE
        
        // FALSE== 1
       
        for(int i=0;i< sieve + 1;i++) 
            prime[i] = false; 
          
        for(int p = 2; p*p <= sieve; p++) 
        { 
           
            if(prime[p] == false) 
            { 
                
                for(int i = p*p; i <= sieve; i += p) 
                    prime[i] = true; 
            } 
        } 
          
       
       
    
    } 
 
///////////////////////////////////////////////////////////////////////////////////


public static String reverse(String input)
{
    String op = "" ;
    
    for(int i = 0; i < input.length() ; i++ )
    {
        op = input.charAt(i)+  op ; 
    }
    
return op ;
}


///////////////////////////////////////////////////////////////////////////////////////


public static void sortD(int[] arr)
{
      sort(arr ,0 , arr.length-1) ;
      
      int i =0 ; int j = arr.length -1  ;
      
      while( i < j)
      {
            int temp = arr[i] ;
            arr[i] =arr[j] ;
            arr[j] = temp ;
            i++ ; j-- ;
      }
      
      
      
      return ;
}

///////////////////////////////////////////////////////////////////////////////////////

public static boolean sameParity(long a  ,long b )
{
      
      if(a%2 == b %2)
      {
            return true  ;
      }
      else{
            return false  ;
      }
}

////////////////////////////////////////////////////////////////////////////////////

public static boolean isPossibleTriangle(int a  ,int b , int c)
{
      if( a + b > c && c+b > a && a +c > b)
      {
            return true  ;
      }
      
      return false  ;
      
      
}

/////////////////////////////////////////////////////////////////////
    
public static int gcd(int a, int b )
{

if(b==0)return a ;

else return gcd(b,a%b) ; 


}

////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c , int d )
{

int temp = lcm(a,b , c) ;


 
 int ans = lcm(temp ,d ) ;

return ans  ;


}

//////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c )
{

int temp = lcm(a,b) ;

int ans =  lcm(temp ,c) ;

return ans  ;


}

////////////////////////////////////////////////////////////////////////////////////////
    
public static int lcm(int a , int b )
{

int gc = gcd(a,b);

return (a*b)/gc ;




}

/////////////////////////////////////////////////////////////////////////////////////////////

static boolean isPrime(int n)
{
      if(n==1)
      {
            return false  ;
      }
      
      boolean ans =  true  ;
      
      for(int i = 2; i*i <= n ;i++)
      {
            if(n% i ==0)
            {
                  ans = false  ;break ;
            }
      }
      
      
      return ans  ;
      
      
}
//////////////////////////////////////////////////////////////////////////////////////////

public static ArrayList<Integer> allFactors(int n)
{   
      ArrayList<Integer> list = new ArrayList<>() ;
      
    for(int i = 1; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n)
              {
                    list.add(i) ;
              }
              else{
                    list.add(i) ;
                    list.add(n/i) ;
                    
              }
          }
    }
      
     return list ; 
      
      
}



/////////////////////////////////////////////////////////////////////////////

public static boolean isPowerOfTwo(int n) 
{ 
    if(n==0) 
    return false; 
  
if(((n ) & (n-1)) == 0 ) return true ;
else return false  ;

} 
   
////////////////////////////////////////////////////////////////////////////////////   
   
    public static int countDigit(long n) 
    { 
        return (int)Math.floor(Math.log10(n) + 1); 
    } 
    
 ///////////////////////////////////////////////////////////////////////////////////////// 
  
   static final int MAXN = 100001; 
       
    
    static int spf[] = new int[MAXN]; 
   
    static void sieve() 
    { 
        spf[1] = 1; 
        for (int i=2; i<MAXN; i++) 
       
           
            spf[i] = i; 
       
       
        for (int i=4; i<MAXN; i+=2) 
            spf[i] = 2; 
       
        for (int i=3; i*i<MAXN; i++) 
        { 
           
            if (spf[i] == i) 
            { 
               
                for (int j=i*i; j<MAXN; j+=i) 
       
                    if (spf[j]==j) 
                        spf[j] = i; 
            } 
        } 
    } 
       
// The above code works well for n upto the order of 10^7.
// Beyond this we will face memory issues.

// Time Complexity: The precomputation for smallest prime factor is done in O(n log log n)
// using sieve.
// Where as in the calculation step we are dividing the number every time by 
// the smallest prime number till it becomes 1.
// So, let’s consider a worst case in which every time the SPF is 2 .
// Therefore will have log n division steps.


// Hence, We can say that our Time Complexity will be O(log n) in worst case.


    static Vector<Integer> getFactorization(int x) 
    { 
        Vector<Integer> ret = new Vector<>(); 
        while (x != 1) 
        { 
            ret.add(spf[x]); 
            x = x / spf[x]; 
        } 
        return ret; 
    } 
       
  
   
public static void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
  public static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

      
         

public static void main (String[] args) throws java.lang.Exception
{
  FastReader scn = new FastReader() ;
int n = scn.nextInt() ;

int[][] arr =new int[n][3] ;

for(int i = 0; i < n ; i++)
{
      arr[i][0] = scn.nextInt() ;
      arr[i][1] = scn.nextInt() ;
      arr[i][2] = scn.nextInt() ;
}


int[][] dp =new int[n][3] ;



// dp[i][0] stores thst if we do activity of type A at ith day,total maximum point till 
//ith day 

// dp[i][1] stores thst if we do activity of type B at ith day,total maximum point till 
//ith day 

// dp[i][2] stores thst if we do activity of type C at ith day,total maximum point till 
//ith day 

//BASE CASE

dp[0][0] = arr[0][0] ;
dp[0][1] = arr[0][1] ;
dp[0][2] = arr[0][2] ;




for(int i = 1 ; i < n ; i++)
{
      dp[i][0] =  arr[i][0] +  Math.max(dp[i-1][1] , dp[i-1][2]) ;
      
      dp[i][1] =  arr[i][1] +  Math.max(dp[i-1][0] , dp[i-1][2]) ;
      
      dp[i][2] =   arr[i][2] +  Math.max(dp[i-1][0] , dp[i-1][1]) ;
      
}


int ans =  Math.max(dp[n-1][0] , Math.max(dp[n-1][1] , dp[n-1][2])) ;


out.println(ans) ;


 
 
out.flush() ;
      
}


}
  
 class Pair implements Comparable<Pair>
{
  int val ;
  
  String str  ;
      
Pair(int val , String str)
{
      this.val = val ;
      this.str = str  ;
}

  public int compareTo(Pair o)
    {
        return this.val - o.val;
    }
 
     
}