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>
typedef  long long ll;

int cmpnum(const void * n1, const void * n2)
{
  if (*(int *)n1 > *(int *)n2)
  {
    return 1;
  }
  else if (*(int *)n1 < *(int *)n2)
  {
    return -1;
  }
  else
  {
    return 0;
  }
}

int main(void)
{

  int x,n;
  int p[200];
  int tmp[200];
  int k=0;
  int i=0;
  scanf("%d %d",&x,&n);
  if(n==0){
    printf("%d\n",x);
    return 0;
  }
  for(int i=0;i<n;i++){
    scanf("%d",&p[i]);
  }

  qsort(p,n,sizeof(int), cmpnum);

  for(int j=1;j<p[n-1];j++){
    if(p[i]!=j){
      tmp[k]=j;
      k++;
    }else{
      i++;
    }
  }


  int min=200;
  int min_ans;
  for(int i=0;i<k;i++){
    //printf("%d\n",x-tmp[i]);
    if(abs(x-tmp[i])<min)
    min_ans=tmp[i];
    min=x-tmp[i];
  }

  printf("%d\n",min_ans);
  return 0;
}