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)

		// --------------------<optimizations>--------------------
#pragma GCC optimize("O3")
//(UNCOMMENT WHEN HAVING LOTS OF RECURSIONS)\
#pragma comment(linker, "/stack:200000000")
//(UNCOMMENT WHEN TRYING TO BRUTEFORCE WITH A LOT OF LOOPS)\
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>

using namespace std;

#define ll long long
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define pii pair<ll,ll>
#define vpii vector<pair<ll,ll> >
#define F first
#define S second
#define ld long double
#define built __builtin_popcountll
#define mst(a,i) memset(a,i,sizeof(a))
#define all(x) x.begin(),x.end()
#define itit(it,a) for(auto it=(a).begin(); it!=(a).end(); it++)
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define repr(i,a,b) for(ll i=a; i>b; i--)
#define reprr(i,a,b) for(ll i=a; i>=b; i--)
#define pi 3.14159265358979323846264338327950288419716939937510582097494459230
ll max3(ll x,ll y,ll z) {return max(max(x,y),z);}
ll min3(ll x,ll y,ll z) {return min(min(x,y),z);}
const ll M=2e5+10,M2=1e6+10,mod=1e9+7,inf=1e17+10;
void add(int& a, int b) {a+=b;if(a>=mod){a-=mod;}}
#define trace1(x)			 	 cerr<<#x<<": "<<x<<endl
#define trace2(x, y)		 	 cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)		 	 cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)	 	 cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)	 cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
int X[] = {0, 1, 0, -1};
int Y[] = {-1, 0, 1, 0};
ll power(ll x,ll n){ll result=1;while(n>0){if(n%2==1)result=(result*x)%mod; x=((x%mod)*(x%mod))%mod;n=n/2;}return result;}


int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	ll n;
	cin >> n;
	ll a[n+1],b[n+1];
	rep(i,0,n+1) cin >> a[i];
	rep(i,0,n) cin >> b[i];
	ll ans=0;
	ll cur=0;
	b[n]=0;
	for(ll i=0; i<=n; i++){
		if(cur>=a[i]){
			ans+=a[i];
			a[i]=0;
		}
		else{
			ans+=cur;
			a[i]-=cur;
		}
		if(b[i]<=a[i]){
			ans+=b[i];
			cur=0;
		}
		else{
			ans+=a[i];
			cur=b[i]-a[i];
		}
	//trace4(cur,a[i],b[i],ans);
	}
	cout << ans;
	return 0;
}
/* The judge is never wrong! Your code is buggy
   Look for:
 * * Read the problem carefully.
 * * Don't forget to sort(), mod, ll!!!!
 * * Initial value = +/- infinity instead of zero!!!
 * * an easier and alternate approach
 * * read the problem statement carefully
 * * if concept is correct and still WA, try with a different implementation
 * * special cases (n=1?)
 * * if you have no idea just guess the appropriate well-known algorithm instead of doing nothing :/
 */