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 sc = new Scanner(System.in);
		int n = sc.nextInt();
		long[] a = new long[n];
		for(int i = 0; i < n; i++) a[i] = sc.nextLong();
		long x = a[0], y = 1;
		for(int i = 0 ; i < n ; i++) {
			x = gcd_lcm.gcd(x, a[i]); // 最大公約数
			y = gcd_lcm.lcm(y, a[i]); // 最小公倍数
		}
		long z = Math.abs(y - x);
		long ans = 0;
		for(int i = 0 ; i < n ; i++) ans += z % a[i];
		System.out.println(ans);
	}
}

class gcd_lcm {
	static long gcd(long a, long b) {
		if(b == 0) return a;
		return gcd(b, a % b);
	}
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}
}