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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
int main(){
   int X,N,p[N],ans,i,j;
   scanf("%d%d",&X,&N);
   int min=abs(p[0]-p[1]);
   for(i=0;i<N;i++){
      scanf("%d",&p[i]);
   }
   for(j=0;j<100;j++){
      int judg=0;
      for(i=0;i<N;i++){
         if(p[i]==j){
            judg++;
            break;
         }
      }
      if(judg==0&&X!=j){
         if(min>abs(X-j)){
            min=abs(X-j);
            ans=j;
         }
      }
   }
   if(N==0){
      ans=X;
   }
   printf("%d\n",ans);
   return 0;
}