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 (1.42.0) (judge verdict: Wrong Answer)

use num::integer::lcm;
use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {
        n: usize,
        a_list: [u128; n]
    }
    let mut m = 1;
    for &a in &a_list {
        m = lcm(m, a);
    }
    m = m - 1;
    println!("{}", a_list.iter().map(|x| m % x).sum::<u128>());
}