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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define int long long
#define ld long double
#define fi first
#define se second
#define all(uiet) uiet.begin(),uiet.end()
#define read(UIET) for(int i = 0; i < n; ++i) cin >> UIET[i]
#define out(UIET) for(int i = 0; i < n; ++i) cout << UIET[i] << " "
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define vpp vector<pair< int, int > >
#define pll pair<int , int >
#define ppll pair < pll , pll >
#define debug(n1) cout << n1 << endl
#define len(a) ((int) (a).size())
#define endl "\n"
#define mod 1000000007
const int INF=(1LL<<60)-1;
bool isPerfectSquare(long double x) {   
    long double sr = sqrt(x); 
    return ((sr - floor(sr)) == 0); 
} 
bool isPrime(int x){
    for(int i = 2; i * i <= x; ++i){
        if(x % i == 0) return 0;
    }
    return 1;
}
int fun(int n){
    if(n & 1) return 3 * n + 1;
    else return n/2;
}
int32_t main(){
    SPEED;
    int n;
    cin >> n;
    int ans = 0;
    while(n--){
        int x;
        cin >> x;
        ans += x - 1;
    }
    cout << ans ;
}