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 <math.h>
#include <stdlib.h>

int min(int x, int y)
{
	if (x < y)
	{
		return x;
	}
	return y;
}

int main()
{
	int x, n;
	scanf("%d %d", &x, &n);

	int p[101] = {0};
	for (int i = 0; i < n; i++)
	{
		int pp;
		scanf("%d", &pp);
		p[pp] = 1;
	}

	int ans = __INT32_MAX__;
	int diff = __INT32_MAX__;
	for (int i = 100; i > 0; i--)
	{
		if (p[i] == 1)
		{
			continue;
		}

		if (abs(x - i) <= diff)
		{
			diff = abs(x - i);
			ans = i;
		}
	}
	printf("%d\n", ans);
	return 0;
}