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: Accepted)

#include<stdio.h>

int main(void)
{
  int x,n;
  scanf("%d %d", &x, &n);

  if(n>0){
    int p[n];
    for(int i=0;i<n;i++){
      scanf("%d", &p[i]);
    }
    int tmp;
    for(int i=0;i<n-1;i++){
      for(int j=i+1;j<n;j++){
        if(p[i] > p[j]){
	  tmp = p[i];
	  p[i] = p[j];
	  p[j] = tmp;
        }
      }
    }

    int j=0;
    int ans;
    int dist=100;
    for(int i=-1;i<=101;i++){
      if(i==p[j]) j++;
      else{
        if(i<x) tmp = x - i;
        else tmp = i - x;

        if(dist>tmp){
	  dist = tmp;
	  ans = i;
        }
      }
    }
  
    printf("%d\n", ans);
  }
  else{
    printf("%d\n", x);
  }

  return 0;
}