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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        //bytesのiterを生成
        .bytes()
        //それをas charでchar型に変換したiterを生成
        .map(|c| c.expect("failed to read char") as char)
        //c.is_whitespace()がfalseになるまで（つまり普通の文字が出るまで）iterを進めたiterを生成
        .skip_while(|c| c.is_whitespace())
        // !c.is_whitespace()がtrue, 即ち（普通の文字の間）iterを進め, 進んだところまでのiterを生成
        .take_while(|c| !c.is_whitespace())
        // それをcollect()してiterからStringを作る
        .collect();
    // parse()で任意の型に変換して返す。
    token.parse().ok().expect("failed to parse token")
}

fn read_one_line_to_vec<T: FromStr>() -> Vec<T> {
    let stdin = stdin();
    let stdin = stdin.lock();
    let v: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c == &'\x0d' || c == &'\x0a')
        .take_while(|c| !(c == &'\x0d' || c == &'\x0a'))
        .collect();
    v.split_whitespace()
        .map(|c| { let string = c.to_string();
            string.parse::<T>().ok().expect("failed to parse token")})
        .collect()
}

fn main() {
    let n: usize = read();

    let a: Vec<i32> = read_one_line_to_vec();
    let b: Vec<i32> = read_one_line_to_vec();

    let mut kurikoshi :i32 = 0;
    let mut count = 0;
    for i in (0..n).rev() {
        if b[i] <= a[i + 1] - kurikoshi {
            // kurikoshi nashi
            count += kurikoshi + b[i];
            kurikoshi = 0;
        } else if b[i] > a[i+1] - kurikoshi {
            // kurikoshi ari
            count += a[i + 1];
            if a[i+1] - kurikoshi <= 0 {
                kurikoshi = b[i];
            } else {
                kurikoshi = b[i] - (a[i+1] - kurikoshi);
            }
        }
    }
    if kurikoshi < a[0] {
        count += kurikoshi;
    } else {
        count += a[0];
    }
    println!("{}", count);
}