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 Ruby (2.3.3) (judge verdict: Wrong Answer)

# N  個の正整数 a1,a2,...,aN が与えられます。
# 非負整数 m に対して、f(m)=(m mod a1)+(m mod a2)+...+(m mod aN) とします。
# ここで、X mod Y は X を Y で割った余りを表します。
# f の最大値を求めてください。

n = gets.to_i
input_a = gets.chomp.split

a = Array.new
n.times do |i|
    a << input_a[i].to_i
end

def sum(m, a, n)
    sum = 0
    n.times do |i|
        sum += (m % a[i])
    end
    return sum
end

patterns = Array.new
3000.times do |i|
    patterns << sum(i+100000, a, n)
end

puts patterns.max