Problem:
Score : 300 points

Problem Statement

You are given N positive integers a_1, a_2, ..., a_N.

For a non-negative integer m, let f(m) = (m\ mod\ a_1) + (m\ mod\ a_2) + ... + (m\ mod\ a_N).

Here, X\ mod\ Y denotes the remainder of the division of X by Y.

Find the maximum value of f.

Constraints

All values in input are integers.

2 \leq N \leq 3000

2 \leq a_i \leq 10^5

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the maximum value of f.

Sample Input 1

3
3 4 6

Sample Output 1

10

f(11) = (11\ mod\ 3) + (11\ mod\ 4) + (11\ mod\ 6) = 10 is the maximum value of f.

Sample Input 2

5
7 46 11 20 11

Sample Output 2

90

Sample Input 3

7
994 518 941 851 647 2 581

Sample Output 3

4527

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io;

fn main() {
    let mut N = String::new();
    let mut astring = String::new();

    io::stdin().read_line(&mut N).expect("fail");
    io::stdin().read_line(&mut astring).expect("fail");

    let N: u32 = N.trim().parse().unwrap();
    let mut a = vec![];
    let mut m: u64 = 1;

    for n in astring.split_whitespace() {
        let n: u64= n.parse().unwrap();
        a.push(n);
        m = m * n;
    }
    //println!("{}{:?}", N, a);
    m = m - 1;
    let mut sum: u64 = 0;
    for n in a.iter() {
        sum = sum + (m % n);
    }

    println!("{}", sum);
}