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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int mod_func(int m, int a);
int LCM(int a, int b);

int main() {
	int ans = 0, max = 1;
	int N;

	scanf("%d", &N);
	int A[N];
	for(int i = 0; i < N; i++) {
		scanf("%d", &A[i]);
	}

//	for(int i = 0; i < N; i++) {
//		max = LCM(max, A[i]);
//	}

//	printf("max = %d\n", max);
	double sum = 0;

	for(int i = 0; i < N-1; i++) {
		max = LCM(A[i], A[i+1]);
		sum += 	(mod_func(max-1, A[i]) + mod_func(max-1, A[i+1])) / 2;
//		printf("%d\n", sum);
	}

	max = LCM(A[0], A[N-1]);
	sum += (mod_func(max-1, A[0]) + mod_func(max-1, A[N-1])) / 2;
//	printf("%d\n", sum);
//	sum /= 2;
	
	printf("%d\n", sum);
	return 0;
}

int mod_func(int m, int a) {
	int ans;
	ans = m % a;
	return ans;
}

int LCM(int a, int b) {
	int x = a * b;

	// 自然数 a > b を確認・入替
	if(a < b) {
		int temp = a;
		a = b;
		b = temp;
	}

	// ユークリッドの互除法
	int r = a % b;
	while(r != 0) {
		a = b;
		b = r;
		r = a % b;
	}

	return x / b;
}