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: Accepted)

#[allow(unused_imports)]
use std::cmp::{min,max};
#[allow(unused_imports)]
use std::collections::BTreeMap;
#[allow(unused_imports)]
use std::ops::*;
#[allow(unused_imports)]
use std::collections::BinaryHeap;

fn main() {
    let n = read_i64() as usize;
    let xs = read_vec_i64();
    let res = xs.iter().fold(0, |r,x| r + x - 1);
    println!("{}", res);
}

#[allow(dead_code)]
fn read_line() -> String {
    let mut ret = String::new();
    std::io::stdin().read_line(&mut ret).ok();
    ret.pop();
    return ret;
}

#[allow(dead_code)]
fn read_i64() -> i64 {
    let ss = read_line();
    return ss.parse::<i64>().unwrap();
}

#[allow(dead_code)]
fn read_vec_i64() -> Vec<i64> {
    let mut res = vec![];
    let ss = read_line();
    for ts in ss.split_whitespace() {
        let x = ts.parse::<i64>().unwrap();
        res.push(x);
    }
    return res;
}

use std::fmt::Display;
#[allow(dead_code)]
fn write_vec<T: Display>(xs: &Vec<T>) {
    if xs.len() == 0 {
        println!("");
        return;
    }
    print!("{}", xs[0]);
    for i in 1..xs.len() {
        print!(" {}", xs[i]);
    }
    println!("");
}