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>

int main(void)
{
	int a[110] = { 0 };
	int x, n;
	int ans1 = 0, kyo = 0;
	int f[110] = { 0 };
	scanf("%d%d", &x, &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		f[a[i]] = 1;
	}

	for (int i = 1; i <= 100; i++) {
		if (i < x) {
			if (f[i] == 0) {
				ans1 = i;
				kyo = x - i;
			}
		}
		else if (x < i) {
			if (ans1 == 0) {
				ans1 = i;
				kyo = i - x;
			}
			else
				if (f[i] == 0 && kyo > i - x)
					ans1 = i;
		}
	}

		if (n == 0)
			printf("%d\n", x);
		else if (ans1)
			printf("%d\n", ans1);

		return 0;

	}