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 x,n,a,b,p[100],i,j,cnt;
	a=0; b=0;
	scanf("%d %d",&x,&n);
	for(i=0;i<n;i++)	scanf("%d",&p[i]);
	for(i=101;i>=x;i--){
		cnt=0;
		for(j=0;j<n;j++){
			if(i!=p[j])	cnt++;
			if(cnt==n)	a=i;
		}
	}
	for(i=0;i<=x;i++){
		cnt=0;
		for(j=0;j<n;j++){
			if(i!=p[j])	cnt++;
			if(cnt==n)	b=i;
		}
	}
	if(n==0)	printf("%d\n",x);
	else{
		if((a+b)/2==x)	printf("%d\n",b);
		else{
			if(a-x<b-x*-1)	printf("%d",a);
			else	printf("%d",b);
		}
	}
	return 0;
}