Problem:
Score : 300 points

Problem Statement

Given are an integer X and an integer sequence of length N: p_1, \ldots, p_N.

Among the integers not contained in the sequence p_1, \ldots, p_N (not necessarily positive), find the integer nearest to X, that is, find the integer whose absolute difference with X is the minimum. If there are multiple such integers, report the smallest such integer.

Constraints

1 \leq X \leq 100

0 \leq N \leq 100

1 \leq p_i \leq 100

p_1, \ldots, p_N are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X N
p_1 ... p_N

Output

Print the answer.

Sample Input 1

6 5
4 7 10 6 5

Sample Output 1

8

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the one nearest to 6 is 8.

Sample Input 2

10 5
4 7 10 6 5

Sample Output 2

9

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the ones nearest to 10 are 9 and 11. We should print the smaller one, 9.

Sample Input 3

100 0

Sample Output 3

100

When N = 0, the second line in the input will be empty. Also, as seen here, X itself can be the answer.

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

#define MAX 200

void input(int *,int n);
int work(int *, int ,int );

int main()
{
	int number[MAX];
	int x, n;

	scanf("%d %d", &x, &n);
	input(number, n);

	printf("%d\n", work(number, x, n));

	return 0;
}

void input(int* number,int n)
{
	int a;

	for (int i = 0; i < MAX; i++) {
		number[i] = i + 1;
	}

	for (int i = 0; i < n; i++) {
		scanf("%d", &a);
		number[a - 1] = 0;
	}

}

int work(int* number, int x,int n)
{
	int r=n;
	int min = MAX;

	for (int i = 0; i < MAX; i++) {
		if (min > abs(number[i] - x)) {
			min = abs(number[i] - x);
			r = i + 1;
		}
	}

	return r;
}