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: Time Limit Exceeded)

import java.io.InputStream;
import java.io.PrintStream;
import java.math.BigDecimal;
import java.util.Scanner;

public class Main {
	InputStream in = System.in;
	PrintStream out = System.out;

	private BigDecimal gcd(BigDecimal a, BigDecimal b) {
		if (b.equals(BigDecimal.ZERO))
			return a;
		return gcd(b, a.remainder(b));
	}

	private BigDecimal lcm(BigDecimal a, BigDecimal b) {
		return a.multiply(b).divide(gcd(a,b));
	}

	private BigDecimal lcm(BigDecimal[] a) {
		BigDecimal ans = a[0];
		for (int i = 1; i < a.length; i++) {
			ans = lcm(ans, a[i]);
		}
		return ans;
	}

	public void _main(String[] args) {
		Scanner sc = new Scanner(in);

		int N = sc.nextInt();

		BigDecimal[] a = new BigDecimal[N];

		for (int i = 0; i < N; i++) {
			a[i] = new BigDecimal(sc.nextInt());
		}

		BigDecimal l = lcm(a);

		l = l.subtract(BigDecimal.ONE);

		BigDecimal z = BigDecimal.ZERO;
		for (int i = 0; i < N; i++) {
			z = z.add( l.remainder(a[i]) );
		}
		out.println(z);

		sc.close();

	}

	public static void main(String[] args) {
		new Main()._main(args);
	}

}