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(void){
  int x,n,p[102],sw[102];
  int i,j;
  
  int min=999,imin;
  
  scanf("%d %d", &x, &n);
  
  if(n==0){
    printf("%d", x);
    return 0;
  }
  
  for(i=0;i<=100;i++){
    sw[i]=999;
  } 
  
  for(i=1;i<=n;i++){
    scanf("%d", &p[i]);
    sw[p[i]]=p[i];
  }
  
  for(i=0;i<=100;i++){
    if(sw[i]==999){
//      printf("%d ", i);
      if(abs(i-x)<min){
        min=abs(i-x);
        imin=i;
      }
    }
  }
  
  printf("%d", imin);
  
  return 0;
}