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 <limits.h> 
 
int main(void) {
    int a[100];int i,o,j;
	for(i=0;i<100;i++){
	    a[i]=i+1;
	}
	int b,c;
	scanf ("%d%d",&b,&c);
	if(i==0){printf("%d",b);}
	int d[c];
    for(i=0;i<c;i++){
        scanf("%d",&d[i]);
    }
    for(i=0;i<100;i++){
        for(o=0;o<c;o++){
        if (a[i]==d[o]){a[i]=0;}
    }}
    for(i=0;i<100;i++){
        for(j=i+1;j<100;j++)
        if(b-a[i]<b-a[j]){
            o=a[i];
            a[i]=a[j];
            a[j]=o;}}
        printf("%d",a[0]);
    return 0;
}