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


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  *********************************************************/
 
#include <bits/stdc++.h>
using namespace std;
 
 
#define ll                          long long
#define ld                          long double
 
#define forn(i,n)                   for(int i=0;i<n;++i)
#define all(v)                      v.begin(), v.end()
#define rall(v)                     v.rbegin(),v.rend()
#define isSubstring(s1,s2)          s1.find(s2) != std::string::npos
	
#define pb                          push_back
#define pf                          push_front
#define mp                          make_pair
#define fr                          first
#define sc                          second
 
#define vpi                         vector<pair<int,int>>
#define vi                          vector<int>
#define sz(a)                       (int)a.size()
 
#define   endl                       "\n"
#define     nl                       cout<<"\n";
#define fastio                       ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 
#define  MN                           INT_MIN
#define  MX                           INT_MAX
#define int long long
bool isPrime(ll n){if(n<2)return false;for(ll x = 2;x*x <=n;x++){if(n%x == 0)return false;}return true;}
bool isPerfectSquare(long double x){long double sr=sqrt(x);return((sr-floor(sr))==0);}
int gcd(int a, int b){if(a==0)return b;if(b==0) return a;if(a==b)return a;if(a>b)return gcd(a-b,b);return gcd(a,b-a);}

int dp[100001];
int32_t main()
{
	int n;
	cin >> n;
	int a[n+1],b[n];
	for(int i = 0;i<n+1;i++)
		cin >> a[i];
	for(int i = 0;i<n;i++)
		cin >> b[i];
	int ans = 0;
	for(int i = 0;i<n;i++)
	{
		ans += min(a[i],b[i]);
		b[i] -= min(a[i],b[i]);
		ans+= min(a[i+1], b[i]);
		a[i+1] -= min(a[i+1],b[i]);
	}
	cout << ans << endl;
}