#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) {
  return n | (1LL << i);
  ;
}
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx4[] = {0, 0, +1, -1};
int dy4[] = {+1, -1, 0, 0};
int dx8[] = {+1, 0, -1, 0, +1, +1, -1, -1};
int dy8[] = {0, +1, 0, -1, +1, -1, +1, -1};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline bool isIntege(double num) { return (num == (int)num); }
inline bool collinear(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2);
}
inline double coDist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
struct edge {
  int p, q, w;
};
bool cmp(edge &a, edge &b) { return a.w < b.w; }
string to_s(int t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
const int mx = 1e5 + 123;
int n;
long long a[mx], b[mx], c[mx];
long long dp[mx][3];
long long solve(int id, int f1) {
  if (id == n - 1) {
    if (f1)
      return b[id];
    else
      return a[id];
  }
  if (dp[id][f1] != -1) return dp[id][f1];
  long long p1 = 0, p2 = 0, p3 = 0;
  if (f1 == 0) {
    p1 = a[id] + solve(id + 1, 1);
    p2 = b[id] + solve(id + 1, 0);
  } else {
    p2 = b[id] + solve(id + 1, 1);
    p3 = c[id] + solve(id + 1, 0);
  }
  do {
    cerr << "p1, p2, p3"
         << " : ";
    faltu(p1, p2, p3);
  } while (0);
  return dp[id][f1] = max(p1, max(p2, p3));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  c[0] = c[n - 1] = -inf;
  memset(dp, -1, sizeof(dp));
  cout << solve(0, 0);
}

### Prompt

Your task is to create a cpp solution to the following problem:

Dima liked the present he got from Inna very much. He liked the present he got from Seryozha even more. 

Dima felt so grateful to Inna about the present that he decided to buy her n hares. Inna was very happy. She lined up the hares in a row, numbered them from 1 to n from left to right and started feeding them with carrots. Inna was determined to feed each hare exactly once. But in what order should she feed them?

Inna noticed that each hare radiates joy when she feeds it. And the joy of the specific hare depends on whether Inna fed its adjacent hares before feeding it. Inna knows how much joy a hare radiates if it eats when either both of his adjacent hares are hungry, or one of the adjacent hares is full (that is, has been fed), or both of the adjacent hares are full. Please note that hares number 1 and n don't have a left and a right-adjacent hare correspondingly, so they can never have two full adjacent hares.

Help Inna maximize the total joy the hares radiate. :)

Input

The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of hares. Then three lines follow, each line has n integers. The first line contains integers a1 a2 ... an. The second line contains b1, b2, ..., bn. The third line contains c1, c2, ..., cn. The following limits are fulfilled: 0 ≤ ai, bi, ci ≤ 105.

Number ai in the first line shows the joy that hare number i gets if his adjacent hares are both hungry. Number bi in the second line shows the joy that hare number i radiates if he has exactly one full adjacent hare. Number сi in the third line shows the joy that hare number i radiates if both his adjacent hares are full.

Output

In a single line, print the maximum possible total joy of the hares Inna can get by feeding them.

Examples

Input

4
1 2 3 4
4 3 2 1
0 1 1 0


Output

13


Input

7
8 5 7 6 1 8 9
2 7 9 5 4 3 1
2 3 3 4 1 1 3


Output

44


Input

3
1 1 1
1 2 1
1 1 1


Output

4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) {
  return n | (1LL << i);
  ;
}
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx4[] = {0, 0, +1, -1};
int dy4[] = {+1, -1, 0, 0};
int dx8[] = {+1, 0, -1, 0, +1, +1, -1, -1};
int dy8[] = {0, +1, 0, -1, +1, -1, +1, -1};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline bool isIntege(double num) { return (num == (int)num); }
inline bool collinear(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2);
}
inline double coDist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
struct edge {
  int p, q, w;
};
bool cmp(edge &a, edge &b) { return a.w < b.w; }
string to_s(int t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
const int mx = 1e5 + 123;
int n;
long long a[mx], b[mx], c[mx];
long long dp[mx][3];
long long solve(int id, int f1) {
  if (id == n - 1) {
    if (f1)
      return b[id];
    else
      return a[id];
  }
  if (dp[id][f1] != -1) return dp[id][f1];
  long long p1 = 0, p2 = 0, p3 = 0;
  if (f1 == 0) {
    p1 = a[id] + solve(id + 1, 1);
    p2 = b[id] + solve(id + 1, 0);
  } else {
    p2 = b[id] + solve(id + 1, 1);
    p3 = c[id] + solve(id + 1, 0);
  }
  do {
    cerr << "p1, p2, p3"
         << " : ";
    faltu(p1, p2, p3);
  } while (0);
  return dp[id][f1] = max(p1, max(p2, p3));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  c[0] = c[n - 1] = -inf;
  memset(dp, -1, sizeof(dp));
  cout << solve(0, 0);
}

```