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: Accepted)

#include <utility>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <numeric>
#define ll long long
#define rep(i,rept) for (ll i = 0; i < rept; i++)
#define Yes(b) std::cout<< (b ? "Yes" : "No")<<std::endl
#define YES(b) std::cout<< (b ? "YES" : "NO")<<std::endl
using namespace std;
//*********************************************************
// 最大公約数（Greatest Common Divisor）を返す。
// 引数に０がある場合は０を返す。
//*********************************************************
int gcd(int m, int n)
{
	// 引数に０がある場合は０を返す
	if ((0 == m) || (0 == n))
		return 0;

	// ユークリッドの方法
	while (m != n)
	{
		if (m > n) m = m - n;
		else         n = n - m;
	}
	return m;
}//gcd
 //*********************************************************
 // 最小公倍数（Least Common Multiple）を返す。
 // 引数に０がある場合は０を返す。
 //*********************************************************
int lcm(int m, int n)
{
	// 引数に０がある場合は０を返す
	if ((0 == m) || (0 == n))
		return 0;

	return ((m / gcd(m, n)) * n); // lcm = m * n / gcd(m,n)
}//lcm
int main() {
	ll n, a[3000], ans = 1, sum = 0;
	cin >> n;
	rep(i, n) {
		cin >> a[i];
		sum += a[i]-1;
	}
	cout << sum;
	return 0;
}