​​传送门​​ Finally, a basketball court has been opened in SIS, so Demid has decided to hold a basketball exercise session. 2⋅n students have come to Demid’s exercise session, and he lined up them into two rows of the same size (there are exactly n people in each row). Students are numbered from 1 to n in each row in order from left to right.

Now Demid wants to choose a team to play basketball. He will choose players from left to right, and the index of each chosen player (excluding the first one taken) will be strictly greater than the index of the previously chosen player. To avoid giving preference to one of the rows, Demid chooses students in such a way that no consecutive chosen students belong to the same row. The first student can be chosen among all 2n students (there are no additional constraints), and a team can consist of any number of students.

Demid thinks, that in order to compose a perfect team, he should choose students in such a way, that the total height of all chosen students is maximum possible. Help Demid to find the maximum possible total height of players in a team he can choose.

Input

The first line of the input contains a single integer n (1≤n≤105) — the number of students in each row.

The second line of the input contains n integers h1,1,h1,2,…,h1,n (1≤h1,i≤109), where h1,i is the height of the i-th student in the first row.

The third line of the input contains n integers h2,1,h2,2,…,h2,n (1≤h2,i≤109), where h2,i is the height of the i-th student in the second row.

Output

Print a single integer — the maximum possible total height of players in a team Demid can choose.

思路：

对于两串序列，看是否选择当前第i个人，判断的依据是：
①若选择第i人，则dp1[i] = dp2[i]+h1[i];
②若不选，则dp1[i] = dp1[i-1];
比较①②的大小即可得出dp[i]的值；
因为有两串，因此用两个dp数组来存放值，最后比较dp1和dp2的值输出即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll h1[100010],h2[100010];
ll dp1[100010],dp2[100010];

int main()
{
  int n;
  cin>>n;
  for(int i = 1; i <= n; i++)
  {
    cin>>h1[i];
  }
  for(int i = 1; i <= n; i++)
  {
    cin>>h2[i];
  }
  for(int i = 1; i <= n; i++)
  {
    dp1[i] = max(dp1[i-1],dp2[i-1]+h1[i]);
    dp2[i] = max(dp2[i-1],dp1[i-1]+h2[i]);
  }
  printf("%lld\n",max(dp1[n],dp2[n]));
  return 0;
}