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 OCaml (4.02.3) (judge verdict: Time Limit Exceeded)

open Scanf
open Printf

let rec scan_list = function
  | 1 -> scanf "%d" (fun k -> [k])
  | n -> scanf "%d " (fun k -> k :: scan_list (n-1))

let rec gcd a b = if b = 0
  then a
  else gcd b (a mod b)

let lcm a b = a * (b / (gcd a b))

let f a m = List.fold_left ( + ) 0 (List.map (fun x -> m mod x) a)

let rec search lst m = function
  | 0 -> m
  | n -> search lst (max m (f lst n)) (n-1)

let () = let lst = scanf "%d\n" scan_list in
  printf "%d\n" (search lst 0 (List.fold_left lcm 1 lst))