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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

#define MIN(a, b) (a < b ? a : b)
#define MAX(a, b) (a > b ? a : b)
#define ABSS(a, b) (a > b ? a - b : b - a)

static uint32_t a[3000];

int main(void) {
	uint32_t N;
	uint32_t fMax = 0;

	scanf("%"PRIu32, &N);
	for (uint32_t idx = 0; idx < N; idx++) {
		scanf(idx != N - 1 ? "%"PRIu32" " : "%"PRIu32, &(a[idx]));
		fMax += a[idx] - 1;
	}

	printf("%"PRIu32, fMax);

	return 0;
}