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()
{
	int x,n,y[100],flg1,a,b,i,cnt1,cnt2;
	scanf("%d %d",&x,&n);
	flg1=0;
	a=x;
	b=x;
	if(n==0){
		printf("%d\n",x);
		// printf("XX\n");
		flg1++;
	}
	else if(flg1==0){
		for(i=0;i<n;i++){
			scanf("%d",&y[i]);
		}
		while(1){
			cnt1=0;
			cnt2=0;
			a=a+1;
			b=b-1;
			for(i=0;i<n;i++){
				if(a==y[i]) {
					cnt1++;
				}
				if(b==y[i]){
					cnt2++;
				}
			}
			if(cnt1==0&&cnt2==0){
				printf("%d\n",b);
				// printf("AA\n");
				break;
			}
			else if(cnt1==0){
				printf("%d\n",a);
				// printf("BB\n");
				break;
			}
			else if(cnt2==0){
				printf("%d\n",b);
				// printf("CC\n");
				break;
			}
		}
	}
	return 0;
}