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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>    // PBDS
#include <ext/pb_ds/tree_policy.hpp>        // PBDS

using namespace std;
using namespace __gnu_pbds;                 // PBDS

/* typedef starts */

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> PBDS_INT;               // PBDS for int
typedef tree<long long, null_type, less<long long>, rb_tree_tag, tree_order_statistics_node_update> PBDS_LL;    // PBDS for long long

/* typedef ends */

/* macro starts */

#define PI acos(-1)
#define EPS 1e-9
#define fastIO ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define sinD(degree) sin((degree * PI) / 180.0)
#define cosD(degree) cos((degree * PI) / 180.0)
#define tanD(degree) tan((degree * PI) / 180.0)
#define cotD(degree) (1.0 / tanD(degree))
#define secD(degree) (1.0 / cosD(degree))
#define cosecD(degree) (1.0 / sinD(degree))
#define asinD(value) (asin(value) * 180.0) / PI
#define acosD(value) (acos(value) * 180.0) / PI
#define atanD(value) (atan(value) * 180.0) / PI
#define forn(i,n) for(int i=0;i<n;i++)
#define value_at_index(x) find_by_order(x)
#define index_of(x) order_of_key(x)
#define pb push_back
#define set0(add) memset(add,0,sizeof(add));

/* macro ends */

/* function starts */

/// calculates n-th (0-based) Gray Code
template<typename dataType>
dataType nthGrayCode(dataType n)
{
    return (n ^ (n >> 1));
}

/// generates all possible subsets for the given set
template<typename dataType>
void generateAllSubset(vector<dataType> &data)
{
    int i, j, n;
    n = data.size();

    for (i = 0; i < (1 << n); i++) {
        for (j = 0; j < n; j++) {
            if (i & (1 << j)) {
                cout << data[j] << " ";
            }
        }
        cout << "\n";
    }
}


void swap(int a, int b){
	int temp;
	temp=a;
	a=b;
	b=temp;
}

int fact(int n){
	if(n<=1)
		return 1;
	else 
	return n*fact(n-1);
}


int n;

int a[100005][3],dp[100005][4];

int func(int ind,int prev)
{
    if(ind==n) return 0;
    if(dp[ind][prev]!=-1)return dp[ind][prev];
    else
	{    int left=-1,right=-1;
		if(prev==1){
		 left=a[ind][1]+func(ind+1,2);
		 right=a[ind][2]+func(ind+1,3);
		}
		else if(prev==2){
		 left=a[ind][0]+func(ind+1,1);
		 right=a[ind][2]+func(ind+1,3);
		}
		else if(prev==3){
           left=a[ind][0]+func(ind+1,1);
		 right=a[ind][1]+func(ind+1,2);
		}
		else if(prev==0){
			int ty;
			ty=a[0][0]+func(ind+1,1);
			int fy=a[0][1]+func(ind+1,2);
			left=max(ty,fy);
			right=a[0][2]+func(ind+1,3);
		}
		return dp[ind][prev]=max(left,right);
	}
	
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif

    fastIO;
    
    cin >> n ;
    for(int i=0;i<n;i++){
     cin >> a[i][0] >> a[i][1] >> a[i][2];
     dp[i][0]=-1;
     dp[i][1]=-1; dp[i][2]=-1; dp[i][3]=-1;
    }
    
    int ans=func(0,0);
    cout << ans;
   
   return 0;
   
   
   	}