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 <stdio.h>
#include <stdlib.h>
int main(){
  int a[200],d[200];
  int b,c,i,j,k,tmp,tmp2,count=0;
  scanf("%d %d",&b,&c);
  for(i=0;i<c;i++){
    scanf("%d",&a[i]);
  }
   for(i=0; i<c; ++i){
    for (j=i+1;j<c;++j){
      if(a[i] > a[j]){
        tmp =  a[i];
        a[i] = a[j];
        a[j] = tmp;
      }
    }
  }
  k=0;
  for(i=-5; i<101; ++i){
    for(j=0; j<c; ++j){
      if(i==a[j])count+=1;
    }
    if(count==0){
      d[k]=i;
      k+=1;
    }
    count=0;
  }
  for(i=0; i<k; ++i){
    for (j=i+1;j<k;++j){
  
      if(abs(b-d[i]) > abs(b-d[j])) {
       tmp2 =  d[i];
        d[i] = d[j];
        d[j] = tmp2;
      }
    }
  } 
  printf("%d",d[0]);
    
  return 0;
  }