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>
using namespace std;
#define FOR(i, x, y) for (int i = x; i < y; i++)
#define FORN(i, x, y) for (int i = x; i <= y; i++)
#define DEC(i, x, y) for (int i = x; i >= y; i--)
#define scan1(X) scanf("%d", &X)
#define scan2(X, Y) scanf("%d%d", &X, &Y)
#define scan3(X, Y, Z) scanf("%d%d%d", &X, &Y, &Z)
#define scan4(X, Y, Z, W) scanf("%d%d%d%d", &X, &Y, &Z, &W)
#define scanl1(X) scanf("%lld", &X);
#define scanl2(X, Y) scanf("%lld%lld", &X, &Y)
#define scanl3(X, Y, Z) scanf("%lld%lld%lld", &X, &Y, &Z)
#define scanl4(X, Y, Z, W) scanf("%lld%lld%lld%lld", &X, &Y, &Z, &W)
#define print1(X) printf("%d\n", X);
#define print2(X, Y) printf("%d %d\n", X, Y);
#define print3(X, Y, Z) printf("%d %d %d\n", X, Y, Z);
#define print4(X, Y, Z, W) printf("%d %d %d %d\n", X, Y, Z, W);
#define printl1(X) printf("%lld\n", X);
#define printl2(X, Y) printf("%lld %lld\n", X, Y);
#define printl3(X, Y, Z) printf("%lld %lld %lld\n", X, Y, Z);
#define printl4(X, Y, Z, W) printf("%lld %lld %lld %lld\n", X, Y, Z, W);
#define max3(X, Y, Z) max(X, max(Y, Z))
#define max4(X, Y, Z, W) max(max(X, Y), max(Z, W))
#define min3(X, Y, Z) min(X, min(Y, Z))
#define min4(X, Y, Z, W) min(min(X, Y), min(Z, W))
#define popcount(X) __builtin_popcount(X)
#define all(x) x.begin(), x.end()
#define size(X) (int)X.size()
#define mp make_pair
#define pb push_back
#define fr first
#define sc second
#define lc(x) 2 * x
#define rc(x) 2 * x + 1
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<ld, ld> pldld;
const long long l_inf = numeric_limits<long long>::max();
const long long l_ninf = numeric_limits<long long>::min();
const int inf = numeric_limits<int>::max();
const int ninf = numeric_limits<int>::min();
const double pi = 1.00 * acos(-1.00);
const double eps = 1e-8;
void cio(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}
ll gcd(ll a, ll b){
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return (a * (b / gcd(a, b))); }
int sq( int x ) {return x * x;}
////////////////////////////////////////////////////////////////////Header
const int maxn = 1e5 + 5;
int n;
int dp[maxn][4];
int w[maxn][4];

int main () {
    scan1(n);
    FORN(i, 1, n) FORN(j, 1, 3) scan1(w[i][j]); 
    FORN(i, 1, n) {
        FORN(j, 1, 3) {
            FORN(k, 1, 3){
                if ( k != j ) dp[i][j] = max(dp[i][j], dp[i-1][k] + w[i][j]);
            }
        }
    }
    int ans = 0;
    FORN(j, 1, 3) ans = max(ans, dp[n][j]);
    print1(ans);
}