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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int N = scan.nextInt();
		long []a = new long[N];
		for(int i = 0; i < N; i++) {
			a[i] = scan.nextLong();
		}
		scan.close();
		long c =  lcm(a[0], a[1]);
		if(N == 2) {
			long ans = (c - 1) % a[0] + (c - 2) % a[1];
			System.out.println(ans);
			System.exit(0);
		}
		for(int i = 2; i < N; i++) {
			c = lcm(c, a[i]);
		}
		long ans1 = 0;
		for(int i = 0; i < N; i++) {
			ans1 += (c - 1) % a[i];
		}

		long ans2 = 0;
		for(int i = 0; i < N; i++) {
			ans2 += (c + 1) % a[i];
		}

		long k = Math.max(ans1, ans2);
		System.out.println(k);


	}
	static long gcd(long a, long b) {
		long v0 = a, v1 = b, v2 = a % b;
		  while (v2 != 0) {
		    v0 = v1;
		    v1 = v2;
		    v2 = v0 % v1;
		  };
		  return v1;
		}

	static	long lcm(long a, long b) {
		  return a * b / gcd(a, b);
		}
}