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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int gcd(long long a, long long b)
{
    return a % b == 0 ? b : gcd(b, a % b);
}

int main()
{
    int n;
    long long num[3000], b, sum;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%lld", &num[i]);
    b = num[0];
    for (int i = 1; i < n; i++) {
        long long c = gcd(b, num[i]);
        b = b * num[i] / c;
    }
    sum = 0;
    for (int i = 0; i < n; i++) {
        sum += (b - 1) % num[i];
    }
    printf("%lld\n", sum);
    return 0;
}