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>

int main(void)
{
        int x,n,i,j,s,c,k=0,min=1000;
        scanf("%d%d",&x,&n);
        int p[n];
        for(i=0;i<n;i++)
                scanf("%d",&p[i]);

        c=100;
        for(i=0;i<=101;i++){
                k=0;
                for(j=0;j<n;j++)
                        if(p[j]==i) k=1;
                if(k!=1){
                        if(x>i) s=x-i;
                        else s=i-x;
                        if(min>s) { min=s; c=i;}
                        else if(min==s && c>i) c=i;
                }
        }
        printf("%d\n",c);
        return 0;
}