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 <math.h>

int main(void) {
	int x,n;
	int p[110];
	int np[110];
	int i,j;
	int flag = 0;
	int ans;
	
	for(i=0;i<105;i++){
		np[i]=i;
	}
	
	scanf("%d",&x);
	//printf("%d\n",x);
	scanf("%d",&n);
	//printf("%d\n",n);
	for(i=1;i<n+1;i++){
		scanf("%d",&p[i]);
		np[p[i]] = -5;
		//printf("%d %d\n",p[i], np[p[i]]);
	}
	
	for(i=1;i<101;i++){
		if(np[i] == x){
			ans = x;
			flag = 1;
		}
	}
	
	if(flag == 0){
		for(i=1;i<101;i++){
			if(np[x+i] == -5 && np[x-i] == -5){
				//printf("NG");
			}
			else if(np[x-i] == -5){
				ans = np[x+i];
				break;
			}
			else {
				ans = np[x-i];
				break;
			}
		}
	}
	
	printf("%d",ans);
	printf("\n");	
	return 0;
}