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>

int main(void){
	int x,n;
	scanf("%d%d",&x,&n);
	if(n==0){
		printf("%d",x);
		return 0;
	}
	int p[n];
	for(int i=0;i<n;i++){
		scanf("%d",&p[i]);
	}
	int seek = 0;
	for(int l=0;l<=51;l++){
			for(int i=0;i<n;i++){
				if(x-l == p[i]){
					seek++;
				}
			}
			if(seek==0){
				printf("%d",x-l);
				return 0;
			}
			seek = 0;
			for(int i=0;i<n;i++){
				if(x+l == p[i]){
					seek++;
				}
			}
			if(seek==0){
				printf("%d",x+l);
				return 0;
			}
			seek=0;
		}
}