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: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte(){
        if(ptr < buflen){
            return true;
        }else{
            ptr = 0;
            try{
                buflen = in.read(buffer);
            }catch(IOException e){
                e.printStackTrace();
            }
            if(buflen <=0){
                return false;
            }
        }
        return true;
    }

    private int readByte(){
        if(hasNextByte())return buffer[ptr++];
        else return -1;
    }

    private static boolean isPrintableChar(int c){
        return 33<=c && c<=126;
    }
    public boolean hasNext(){
        while(hasNextByte() && !isPrintableChar(buffer[ptr]))ptr++;
        return hasNextByte();
    }

    public String next(){
        if(!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)){
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong(){
        if(!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        if(b < '0' || '9' < b){
            throw new NumberFormatException();
        }
        while(true){
            if('0' <= b && b<='9'){
                n*=10;
                n+=b-'0';
            }else if(b==-1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt(){
        long nl = nextLong();
        if(nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDoutble(){return Double.parseDouble(next());}
}

public class Main {
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int n = fs.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        int[] c = new int[n];
        for(int i=0;i<n;i++){
            a[i] = fs.nextInt();
            b[i] = fs.nextInt();
            c[i] = fs.nextInt();
        }

        int[][] dp = new int[n][3];
        dp[0][0] = a[0];
        dp[0][1] = b[0];
        dp[0][2] = c[0];
        for(int i=1;i<n;i++){
            dp[i][0] = Math.max(dp[i-1][1],dp[i-1][2])+a[i];
            dp[i][1] = Math.max(dp[i-1][0],dp[i-1][2])+b[i];
            dp[i][2] = Math.max(dp[i-1][0],dp[i-1][1])+c[i];
        }
        System.out.println(Math.max(Math.max(dp[n-1][0],dp[n-1][1]),dp[n-1][2]));
    }
}