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>
#include <cstdio>
using namespace std;
#define ll  long long
#define PB push_back
#define ld long double
#define ff first
#define ss second
# define st(v) (v).begin(),(v).end()
#define   pr pair<int,int>
using namespace std::chrono;
const int N = 1e5+4 , M=N,inf=(int)1e9;
const ll mod=1e9+7;
int fastAbs(int n) { return (n ^ (n >> 31)) - (n >> 31); }
ll multiply(ll a, ll b){ return ((a % mod) * (b % mod)) % mod; }
ll add(ll a, ll b) {     return ((a % mod) + (b % mod)) % mod; }
ll sub(ll a, ll b) { return ((a%mod) - (b % mod)+ mod) % mod ; }
map< pair<int,int> , int > mp;
const int dx[4] = {1,0,0,-1};
const int dy[4] = {0,-1,1,0};
void solve(){
   int n ;
   cin >> n ;
   vector < int > a(n);
   ll ans = 0 ;
   for( int i = 0 ; i < n ; i++) cin >> a[i] , ans+=a[i]-1;
   cout << ans ;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int q=1;
    //cin >> q;
    while(q--)
        solve();
}