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: Accepted)

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cfloat>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <sstream>
#include <stack>
#include <time.h>
#include <vector>
#include <complex>
#include <map>
#include <set>
#include <iomanip>
#include <math.h>
#include <stdlib.h>
#include <list>
#include <utility>
#include <memory>
#include <cstring>
#include <fstream>
#include <numeric>
#include <assert.h>
#include <bits/stdc++.h>
using namespace std;

int a[3003];

int main()
{
  int n,sum=0;
  cin>>n;
  for (int i=0;i<n;i++)
  {
  	cin>>a[i];
  	sum+=a[i];
  }
  cout<<sum-n<<endl;
  return 0;
}