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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, double> Pid;
typedef pair<double, int> Pdi;
typedef pair<ll, int> Pl;
const double PI = 3.1415926535897932;   // acos(-1)
const double EPS = 1e-15;
const int INF = 1001001001;
const ll mod = 1e+9 + 7;

#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define chadd(x, y) x = (x + y) % mod

ll lcm(ll a, ll b){
    return a / __gcd(a, b) * b;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<ll> a(n);
    ll l = 1;
    for(int i = 0; i < n; ++i){
        cin >> a[i];
        l = lcm(l, a[i]);
    }
    --l;
    ll res = 0;
    for(int i = 0; i < n; ++i)  res += l % a[i];
    cout << res << endl;
}