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(){
   int x,p[100],i,n,u,d;

   scanf("%d %d",&x,&n);
   for (i=0;i<n;i++) scanf("%d",&p[i]);
   for (i=0;i<n;i++)
      if (p[i]==x) break;
   if (i==n) {
      printf("%d\n",x);
      return 0;
   }
   d=x-1;u=x+1;
   for (;n>0;) {
      for (i=0;i<n;i++)
         if (p[i]==d) break;
      if (i==n) {
         printf("%d\n",d);
         return 0;
      } else {
         p[i]=p[n-1];
         n--;
      }
      d--;
      for (i=0;i<n;i++)
         if (p[i]==u) break;
      if (i==n) {
         printf("%d\n",u);
         return 0;
      } else {
         p[i]=p[n-1];
         n--;
      }
      u++;
   }
   printf("%d\n",d);
   return 0;
}