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: Accepted)

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

int main()
{
	int x, n;
	while (scanf("%d%d", &x, &n) != EOF) {
		if (n == 0) {
			printf("%d\n", x);
			continue;
		}
		int i, p;
		int f[105];
		memset(f, 0, sizeof(f));
		for (i = 0; i < n; i ++) {
			scanf("%d", &p);
			f[p] = 1;
		}
		
		int a1, a2;
		int mina = 9999;
		a1 = 0;
		for (i = 1; i <=101; i ++) {
			if (f[i] == 0) {
				int diff = abs(i - x);
				if (diff < mina) {
					a1 = i;
					mina = diff;
				}
			}
		}
		
		if (abs(a1 - x) < abs(0 - x)) {
			printf("%d\n", a1);
		} else {
			printf("%d\n", 0);
		}
	}
	return (0);
}