Problem:
Score : 300 points

Problem Statement

There are N+1 towns. The i-th town is being attacked by A_i monsters.

We have N heroes. The i-th hero can defeat monsters attacking the i-th or (i+1)-th town, for a total of at most B_i monsters.

What is the maximum total number of monsters the heroes can cooperate to defeat?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N+1}
B_1 B_2 ... B_N

Output

Print the maximum total number of monsters the heroes can defeat.

Sample Input 1

2
3 5 2
4 5

Sample Output 1

9

If the heroes choose the monsters to defeat as follows, they can defeat nine monsters in total, which is the maximum result.

The first hero defeats two monsters attacking the first town and two monsters attacking the second town.

The second hero defeats three monsters attacking the second town and two monsters attacking the third town.

Sample Input 2

3
5 6 3 8
5 100 8

Sample Output 2

22

Sample Input 3

2
100 1 1
1 100

Sample Output 3

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
//登録しよう！→ https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
using namespace std;
using ll = long long;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}//a,bの最大公約数(gcd)を求める
ll lcm(ll a,ll b){return abs(a*b)/gcd(a,b);}//a,bの最小公倍数(lcm)を求める
vector<ll> enum_div(ll n){vector<ll> ret;for(int i=1 ; i*i<=n ; i++){if(n%i == 0){ret.push_back(i);if(i!=1 && i*i!=n)ret.push_back(n/i);}}ret.push_back(n);return ret;}
//↑nの約数を求める
vector<bool> IsPrime; void sieve(size_t max){if(max+1 > IsPrime.size())IsPrime.resize(max+1,true);IsPrime[0] = false;IsPrime[1] = false;for(size_t i=2; i*i<=max; ++i)if(IsPrime[i])for(size_t j=2; i*j<=max; ++j)IsPrime[i*j] = false;}
//↑エラトステネスの篩で素数を求める
#define roundup(divisor,dividend) (divisor + (dividend - 1)) / dividend //切り上げ割り算
#define all(x) (x).begin(),(x).end() //xの初めから終わりまでのポインタ
#define size_t ll //size_tは自動でllに変換される
#define pb(x) push_back(x)
#define gre greater //sortを降順にする
#define pri_queue priority_queue //優先度付きキュー
#define syo(x) fixed << setprecision(x) //iostreamで小数をx桁表示

void solve(long long N, std::vector<long long> A, std::vector<long long> B){
  ll ans = 0;
  for (int i = 0; i < N; i++) {
    ans += min(A[i],B[i]);
    if(A[i] < B[i]){
      ans += min(B[i]-A[i] , A[i+1]);
      A[i+1] -= min(B[i]-A[i] , A[i+1]);
    }
  }
  cout << ans << endl;
}

// Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    long long N;
    scanf("%lld",&N);
    std::vector<long long> A(N+1);
    for(int i = 0 ; i < N+1 ; i++){
        scanf("%lld",&A[i]);
    }
    std::vector<long long> B(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%lld",&B[i]);
    }
    solve(N, std::move(A), std::move(B));
}