Problem:
Score : 200 points

Problem Statement

There are N gems. The value of the i-th gem is V_i.

You will choose some of these gems, possibly all or none, and get them.

However, you need to pay a cost of C_i to get the i-th gem.

Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.

Find the maximum possible value of X-Y.

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq C_i, V_i \leq 50

Input

Input is given from Standard Input in the following format:

N
V_1 V_2 ... V_N
C_1 C_2 ... C_N

Output

Print the maximum possible value of X-Y.

Sample Input 1

3
10 2 5
6 3 4

Sample Output 1

5

If we choose the first and third gems, X = 10 + 5 = 15 and Y = 6 + 4 = 10.
We have X-Y = 5 here, which is the maximum possible value.

Sample Input 2

4
13 21 6 19
11 30 6 15

Sample Output 2

6

Sample Input 3

1
1
50

Sample Output 3

0

Solution in Rust (1.15.1) (judge verdict: Accepted)



fn main() {
    let n: usize = read();
    let v: Vec<isize> = read_vec();
    let c: Vec<isize> = read_vec();

    let mut d: Vec<isize> = Vec::new();
    for i in 0..n {
        d.push(v[i] - c[i]);
    }
    
    let mut ans = 0;
    for i in d {
        if i > 0 {
            ans += i;
        }
    }
    println!("{}", ans);
}

// ########################################################################################### //

// 標準入力から一行を読み取り、指定の型に変換する関数
pub fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

// 標準入力から一行を読み取り、空白文字で分割し、各要素を指定の型に変換する関数
pub fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

// 標準入力からn行を読み取り、各行を空白文字で分割し、各要素を指定の型に変換する関数
pub fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

// 10進数での各桁の和
pub fn sum_dights(x: isize) -> isize {
    if x % 10 == 0 {
        return 0
    }
    let m = x % 10;
    m + sum_dights((x - m) / 10)
}