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: Runtime Error)

#include<stdio.h>

int main(){
	int x,n;
	scanf("%d %d",&x,&n);
	if(n == 0){
		printf("%d",x);
		return 0;
	}
	int i,p[n + 1],tmp,min,max;
	for(i = 0; i < n; i++){
		scanf("%d",&tmp);
		p[tmp] = 1;
	}
	min = x;
	max = x;
	while(1){
		/*if(n == 0){
			printf("%d",x);
			break;
		}*/
		if(p[min] != 1 || p[max] != 1){
			if(p[min] != 1){
				printf("%d",min);
				break;
			} else {
				printf("%d",max);
				break;
			}
		}
		min--;
		max++;
		if(min == 0){
			printf("%d",min);
			break;
		}
		if(n + 1 < max){
			max = n + 1;
		}
	}
	/*if(n == 0){
		printf("%d",x);
	}*/

	return 0;
}