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)

//abc103_c.cpp
//Sat Jul 21 20:50:52 2018

#include <iostream>
#include <string>
#include <queue>
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <math.h>
#define INF 100000000
using namespace std;
using ll=long long;
typedef pair<int,int> P;

long long gcd (long long a, long long b){ //最大公倍数
	if (b == 0){
		return a;
	}
	return gcd(b, a%b);
}

long long lcm (long long a, long long b){ //最大公約数
	long long temp = gcd(a,b);
	return (a/temp)*b;
}

int main(){
	int n;
	cin >> n;

	ll a[n];
	ll ans = 1;
	for (int i=0;i<n;i++){
		cin >> a[i];
		ans = lcm(ans,a[i]);
	}

	ans--;
	ll rans = 0;
	for (int i=0;i<n;i++){
		rans += ans%a[i];
	}

	cout << rans << endl;
}