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<iostream>
#include<cstring>
//#include<bits/stdc++.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<cstdio>
#include<map>
#include<set>
#define  si(a)       scanf("%d",&a)
#define  sl(a)       scanf("%lld",&a)
#define  sii(a,b)    scanf("%d%d",&a,&b)
#define  sll(a,b)    scanf("%lld%lld",&a,&b)
#define  queues      priority_queue
#define mod 998244353
#define mem(a)  memset(a,0,sizeof(a));
#define def(a) ((a)&(-a))
#define fi  first
#define se  second
#define mp  make_pair
#define  pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
typedef long long ll;
//priority_queue<ll,vector<ll >,greater<ll > >q;
const int INF=0x3f3f3f3f;
//const double E=exp(1);
//const double PI=acos(-1);
using namespace std;
//__builtin_popcount
//priority_queue
const ll MAX=100000;
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
ll kc(ll a,ll b)
{
    ll c=1;
    while(b)
    {
        if(b&1)c=(c*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return c;
}
bool cmp(double x,double y)
{
    return x>y;
}
ll F[50003*3];
ll Find(ll a)
{
    return F[a]<0?a:F[a]=Find(F[a]);
}
ll same(ll a,ll b)
{
    if(Find(a)==Find(b))return 1;
    return 0;
}
ll up1(ll a,ll b)
{
    ll x,y;
    x=Find(a);
    y=Find(b);
    if(x!=y)
        F[x]=y;
    return 0;
}
int read() {
    int x=0,f=1;char s=getchar();
    for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
    for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
    return x*f;
}ll a[100003];
ll  b[100003];
int main()
{
   // ios::sync_with_stdio(false);
    int t;
  cin>>t;
   for(int i=1;i<=t+1;i++)
    cin>>a[i];
    ll sum=0;
   for(int i=1;i<=t;i++)
   {
       cin>>b[i];
   }
   for(int i=1;i<=t;i++)
   { ll z=min(a[i],b[i]);
       sum+=z;b[i]-=z;
        z=min(a[i+1],b[i]);
       sum+=z;a[i+1]-=z;

   }cout<<sum;
}