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<bits/stdc++.h>
using namespace std;
 
#define P(p) cout<<(p)<<endl
#define rep(i,m,n) for(int i = (m); i < (int)(n); i++)
#define rrep(i,m,n) for(int i=(int)(m); i>=(int)(n); i--)
#define vsort(v) sort(v.begin(), v.end());
#define rvsort(v) sort(v.begin(), v.end(),greater<int>());
#define YES cout<<"YES"<< endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl  
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
#define ret return
#define lb(v,n) lower_bound(v.begin(),v.end(),n)
#define ub(v,n) upper_bound(v.begin(),v.end(),n)
#define mae(v) max_element(v.begin(),v.end())
#define mie(v) min_element(v.begin(),v.end())
#define INF 0x7FFFFFFF
#define mod 1000000007
typedef long long ll;
////////////////////////////////////////////////////////////

int main(){
	int n;
	cin >> n;

	ll ans=0;

	rep(i,0,n){
		int a;
		cin >> a;
		ans += a-1;
	}

	cout << ans << endl;
	ret 0;
}