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>
int main(void)
{
  int x,n,a,b;
  int iscont_a = 0,iscont_b=0;
  scanf("%d %d",&x,&n);
  
  if(n == 0){
    printf("%d\n",x);
    return 0;
  }
  
  a = x-1;
  b = x+1;
  
  int p[n];
  for(int i=0; i<n; i++)
    scanf("%d",&p[i]);
  
  while(1){
    for(int i=0; i<n; i++){
      if(a == p[i])
        iscont_a = 1;
      if(b == p[i])
        iscont_b = 1;
    }
    if(!iscont_a){
      printf("%d\n",a);
      return 0;
    }
    else if(!iscont_b)
    {
      printf("%d\n",b);
      return 0;
    }
    a--;
    b++;
    iscont_a=0;
    iscont_b=0;
  }
  return 0;
}