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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define repl(i,x,n) for(long long i=(long long)(x);i<(long long)(n);i++)
#define rep(i,x,n) for(long i=long(x);i<long(n);i++)
/*#define cin fin
#define cout fout
ifstream cin("input.in");ofstream cout("output.out");*/
long long HCF(long long a,long long b){
	if(a==0) return b;
	return HCF(b%a,a);
}
long long LCM(long long a,long long b){
	return (a*b)/HCF(a,b);
}
int main() 
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    long long n,x,a[10009];
    cin>>n>>a[0];
    x=a[0];
    rep(i,1,n){
    	cin>>a[i];
    	x=LCM(x,a[i]);
    }
    //cout<<x<<"\n";
    long long p=0;
    x--;
    for(long i=0;i<n;i++)
   		p+=x%a[i];
   	cout<<p<<"\n";
}