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: Accepted)

#include <stdio.h>

#define swap(X, Y) {long long temp = (X); (X) = (Y); (Y) = temp;}

long long gcd(long long n, long long m);
long long lcm(long long n, long long m);

int main()
{
    long long n;
    long long a[3005];
    long long ans = 0;
    int i, j;

    scanf("%lld", &n);
    for (i = 0; i < n; i++) {
        scanf("%lld", &a[i]);
        ans += a[i] - 1;
    }

    printf("%lld\n", ans);

    return 0;
}

long long gcd(long long n, long long m)
{
    if (m < n) {
        swap(n, m);
    }

    long long r;
     while (n != 0) {
        r = m % n;
        m = n;
        n = r;
    }

    return m;
}
        
long long lcm(long long n, long long m)
{
    return n * m / gcd(n, m);
}