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;
  int N;
  int p[100];

  scanf("%d", &X);
  scanf("%d", &N);

  for(int i = 0;i<N;i++){
      scanf("%d", &p[i]);
  }

  int absolute = 0;
  int ans=0;
  while(X != absolute){
    for(int i = 0;i<N;i++){
        if(X-absolute == p[i]){
          break;
        }

        if(i == N-1){
          ans = X-absolute;
          printf("%d\n",ans);
          return 0;
        }
    }
    for(int j = 0;j<N;j++){
      if(X+absolute == p[j]){
        break;
      }

      if(j == N-1){
        ans = X+absolute;
        printf("%d\n",ans);
        return 0;
      }

    }
      absolute++;
  }

        printf("%d\n",X);

  return 0;
}