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: Time Limit Exceeded)

#include <iostream>                                                                                                                                                                                                                                                                       
                                                                                                                                                                                                                                                                                          
int main(void) {                                                                                                                                                                                                                                                                          
  int N;                                                                                                                                                                                                                                                                                  
  std::cin >> N;                                                                                                                                                                                                                                                                          
  int a[N];                                                                                                                                                                                                                                                                               
  for(int i = 0; i < N; i++) {                                                                                                                                                                                                                                                            
    std::cin >> a[i];                                                                                                                                                                                                                                                                     
  }                                                                                                                                                                                                                                                                                       
  int max = 0;                                                                                                                                                                                                                                                                            
  int f=0;                                                                                                                                                                                                                                                                                
  int ans = 0;                                                                                                                                                                                                                                                                            
  long int N_max = 10000000000;                                                                                                                                                                                                                                                           
  for(long int j = 0; j < N_max; j++) {                                                                                                                                                                                                                                                   
    f = 0;                                                                                                                                                                                                                                                                                
    for(int i = 0; i < N; i++) {                                                                                                                                                                                                                                                          
      f = f + j % a[i];                                                                                                                                                                                                                                                                   
    }                                                                                                                                                                                                                                                                                     
    if(f > max){                                                                                                                                                                                                                                                                          
      max = f;                                                                                                                                                                                                                                                                            
      ans = j;                                                                                                                                                                                                                                                                            
    }                                                                                                                                                                                                                                                                                     
  }                                                                                                                                                                                                                                                                                       
  std::cout << max << std::endl;                                                                                                                                                                                                                                                          
  return 0;                                                                                                                                                                                                                                                                               
}