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>
using namespace std;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define ll long long int
#define vi vector<int>
#define vl vector<ll>
#define v2i vector<vector<int>>
#define v2l vector<vector<ll>>
#define ppi pair<int,int>
#define vpi vector<ppi>
#define ppl pair<ll,ll>
#define all(x) x.begin(),x.end()
#define ii int,int 
#define ff first    
#define ss second
#define forn(i,a,b) for(int i=a;i<b;i++)
#define forr(i,a,b) for(int i=a;i>=b;i--)
#define forv(i,m) for(auto i : m)
#define imx INT_MAX
#define imn INT_MIN
#define inf LONG_MAX
#define minf LONG_MIN
#define endl "\n"
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define sze size()
#define rvs reverse 
#define itr iterator 
#define pre cout<<fixed<<setprecision(10);
#define umap unordered_map
#define uset unordered_set

/*
#define MODM(a,b,md) ((a%md)*(b%md))%md
#define MODA(a,b,md) ((a%md)+(b%md))%md
#define ull unsigned long long int
#define si short int
*/

/*
bool compare(int a,int b) {return a > b;}
bool compare1(ppi a,ppi b) {return a.ff > b.ff;}
bool compare2(ppi a,ppi b) {return a.ss < b.ss;}
ll Npower(ll a,ll b) {ll product = 1; while(b){if(b&1){product = product*a;}a = a*a;b = b>>1;} return product;}
ll power(ll a,ll b,ll md) {ll product = 1; while(b){if(b&1){product = MODM(product,a,md);}a = MODM(a,a,md);b = b>>1;} return product%md;}
ll GCD(ll a,ll b) {return b==0 ? a:GCD(b,a%b);}
ll LCM(ll a,ll b) {return (a/GCD(a,b))*b; }
*/

int main()
{
    fast


        
    int n;
    cin>>n;

    vl v1(n+1),v(n);    

    forn(i,0,n+1)
    {
        cin>>v1[i];
    }


    forn(i,0,n)
    {
        cin>>v[i];
    }

    ll ans = 0;
    forn(i,0,n)
    {
        if(v1[i] <= v[i])
        {
            ans += v1[i];
            v[i] -= v1[i];
            v1[i] = 0;
        }
        else
        {
            ans += v[i];
            v1[i] -= v[i];
            v[i] = 0;
        }
        // cout<<ans<<" ";
        if(v1[i+1] <= v[i])
        {
            ans += v1[i+1];
            v[i] -= v1[i+1];
            v1[i+1] = 0;
        }
        else
        {
            ans += v[i];
            v1[i+1] -= v[i];
            v[i] = 0;
        }

        // cout<<ans<<" ";
    }

    cout<<ans;







    return 0;
}