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>

using namespace std;

typedef long long ll;

const ll LongINF=1e13+7;
const int INF=1e9+7;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};

template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}
template <class T, class Y>T GCD(T a, Y b){if(a<b){int c=a;a=b;b=c;}if (a % b == 0){return b;}return GCD(b, a % b);}
void clear(queue<pair<int, int>> &q){queue<pair<int, int>> empty;swap(q, empty);} //queueの中身の型は適時変更を忘れない

#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define rv reverse

//最小公倍数-1のmod?
//感覚で解くのよくなかった

int main(){
    int N;
    ll product=1;
    cin>>N;
    vector<int> A(N);
    rep(i,N){
        cin>>A[i];
        product*=A[i];
    }
    product--;


    int ANS=0;

    rep(i,N){
        ANS+=product%A[i];
    }

    cout<<ANS<<endl;

    return 0;
}

//a=97,z=122