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<bits/stdc++.h>
#include <boost/math/common_factor_rt.hpp>
const int INF = 1e9;
const int MOD = 1e9+7;
using LL = long long;
const LL LINF = 1e18;
using namespace std;
#define COUT(v) cout<<(v)<<endl
#define CIN(n)  int(n);cin >> (n)
#define LCIN(n) LL(n);cin >> (n)
#define SCIN(n) string(n);cin >> (n)
#define YES(n) cout<<((n)? "YES" : "NO")<<endl
#define Yes(n) cout<<((n)? "Yes" : "No")<<endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) <<endl

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)

#define FOREACH(x,a) for(auto& (x) : (a) )

#define ALL(obj) (obj).begin(),(obj).end()

#define P pair<LL,LL>
#define I vector<int>
#define S set<int>
#define pb(v) push_back(v)
#define V vector
#define rt return
#define rmsame(a) sort(ALL(a)),a.erase(unique(ALL(a)), a.end())

typedef string::const_iterator State;
class PalseError {};
class Edge{
public:
    LL from,to,value;
    Edge(LL a,LL b,LL c){
        from = a;
        to = b;
        value = c;
    }
    Edge(LL a,LL b){
        from = a;
        to = b;
    }
};

int main(){
    CIN(N);
    V<LL> vec;
    LL ins = 1;
    REP(a,N){
        LCIN(b);
        vec.pb(b);
        ins = boost::math::lcm(b,ins);
    }
    ins -= 1;
    LL ans = 0;
    REP(a,N){
        ans += ins % vec.at(a);
    }
    COUT(ans);
    return 0;
}