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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static void Main()
    {
        Console.WriteLine(Solve());
        //Solve();
        return;
    }

    public static string Solve()
    {
        int N = ReadInt();
        int[] A = ReadInts();
        int[] B = ReadInts();
        long ans = 0;
        for (int i = 0; i < N; i++)
        {
            if (B[i] >= A[i])
            {
                ans += A[i];
                B[i] -= A[i];
                if (B[i] >= A[i + 1])
                {
                    ans += A[i + 1];
                    A[i + 1] = 0;
                }
                else
                {
                    ans += B[i];
                    A[i + 1] -= B[i];                    
                }
            }
            else
            {
                ans += B[i];
            }
        }
        return ans.ToString();
    }

    public static long Lcm(long a, long b)
    {
        return a * b / Gcd(a, b);
    }

    public static long Gcd(long a, long b)
    {
        if (a < b)
            // 引数を入替えて自分を呼び出す
            return Gcd(b, a);
        while (b != 0)
        {
            var remainder = a % b;
            a = b;
            b = remainder;
        }
        return a;
    }

    private static string Read() { return Console.ReadLine(); }
    private static int ReadInt() { return int.Parse(Read()); }
    private static long ReadLong() { return long.Parse(Read()); }
    private static double ReadDouble() { return double.Parse(Read()); }
    private static int[] ReadInts() { return Array.ConvertAll(Read().Split(), int.Parse); }
    private static long[] ReadLongs() { return Array.ConvertAll(Read().Split(), long.Parse); }
    private static double[] ReadDoubles() { return Array.ConvertAll(Read().Split(), double.Parse); }
}