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>
#include <string.h>
int main() {
  int i;
  long long int n,m,r,x,tmp,fmax=0,a[3010],b[3010];
  scanf("%lld",&n );
  for(i=0;i<n;i++){
  scanf("%lld",&a[i]);
  b[i]=a[i];
}
for(i=0;i<n-1;i++){
  if(a[i]<a[i+1]){
    tmp=a[i];
    a[i]=a[i+1];
    a[i+1]=tmp;
  }
  x=a[i]*a[i+1];
  r=a[i]%a[i+1];
  while(r!=0){
    a[i]=a[i+1];
    a[i+1]=r;
    r=a[i]%a[i+1];
  }
  a[i+1]=x/a[i+1];
}
  m=a[n-1]-1;  //割られる数
  for(i=0;i<n;i++)
  fmax+=m%b[i];
  printf("%lld\n",fmax );
}