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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.PrintStream;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = null;
		PrintStream out = null;
		try {
			sc = new Scanner(System.in);
			out = System.out;
			exec(sc, out);
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		} finally {
			if(out != null) {
				out.close();
			}
			if(sc != null) {
				sc.close();
			}
		}
		System.exit(0);
	}
	public static void exec(final Scanner sc, final PrintStream out) throws Exception{
		final int N = sc.nextInt();
		final long[] A = new long[N+1];
		final long[] B = new long[N];
		for(int i=0;i<N+1;i++) {
			A[i] = sc.nextLong();
		}
		for(int i=0;i<N;i++) {
			B[i] = sc.nextLong();
		}
		long left = 0;
		long sum = 0;
		for(int i=0;i<N;i++) {
			if(A[i]>B[i]+left) {
				// 全消費しても倒しきれない。
				sum+=(B[i]+left);
				left = 0;
			}else {
				// リソース余り(全部倒せる)
				sum+=A[i];
				if(A[i]>left) {
					// 積み立て全消費
					// 積み立てからのあまりを引き継ぐ。
					left = (B[i])-(A[i]-left);
				}else {
					// 積み立て消失+全部再積み立て
					left = B[i];
				}
			}
		}
		if(A[N]>left) {
			sum += left;
		}else {
			sum += A[N];
		}
		out.println(sum);
	}
}