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 <iostream>
using namespace std;

long koubai(long a,int b){
    
    int i=2;
    long c=a;
    while(c%b!=0){
        c=a*i;
        i++;
    }
   // cout<<c<<endl;
    return c;
}


int main(){
    int n;
    cin>>n;
    long suu[n];
    long koubaisuu=1;
    int ans=0;
    for(int i=0;i<n;i++){
        int a;
        cin>>a;
        suu[i]=a;
        koubaisuu=koubai(koubaisuu,a);
    }
    
    
    for(int i=0;i<n;i++){
        ans+=(koubaisuu-1)%suu[i];
    }
    cout<<ans;





return 0;
}