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<string.h>
 static long Gcd(long m, long n)
{
	long temp;
	while (m % n != 0)
	{
		temp = n;
		n = m % n;
		m = temp;
	}
	return n;
}


 static long Lcm(long m, long n)
{
	return m * n / Gcd(m,n); 
}


int main(void){
   int N,a[4000],big=0,mod,x,tmp,r,w,b,i=0;
   long A[4000],m;
   scanf("%d",&N);
   for(i=0;i<N;i++){
      scanf("%d",&a[i]);
      A[i]=a[i];
   }
   for(i=0;i<N-1;i++){
      A[i+1]=Lcm(A[i],A[i+1]);
   }
   m=A[N-1];

     for(i=0;i<N;i++){
      mod+=(m-1)%a[i];
   }

   printf("%d",mod);

   return 0;
}