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 (Clang 10.0.0) (judge verdict: Runtime Error)

#include<stdio.h>
int main(){
  int X,N,a,i,p[N];
  
  scanf("%d",&X);
  scanf("%d",&N);
  
  for(i=1;i<=N;i++)
    scanf("%d",&p[i]);
  
  
  for(i=1;i<=N;i++){
    a=p[i];
    if(X-p[i]>X-p[i+1]){
      a=p[i+1];
    }
      else if(X-p[i]==X-p[i+1]){
        if(p[i]>p[i+1]){
          a=p[i+1];
        }
      }
  }
  
  printf("%d",a);
    return 0;
  }