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: Time Limit Exceeded)

#include <stdio.h>
#include <limits.h>
int func(int,int*);
int main(){
	int n;
	scanf("%d",&n);
	int a[n];
	int i;
	int* p = a;
	for(i=1;i<=n;i++){
           scanf("%d",p);
	   p++;
	}
	printf("%d\n",func(n,a));
}
int func(int size,int* p){
	int i,n;
	int max = 0;
	int modsum = 0;
	for(i=1;i<INT_MAX;i++){
		modsum = 0;
		for(n=0;n<size;n++){
			modsum += i%p[n];
		}
		if(modsum > max) max = modsum;
		if(modsum==0) break;
	}
	return max;
}