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)

/* abc_170_c */

#include<stdio.h>
#include<stdlib.h>
 
int main()
{
  int X,N;
  int p[100];
  int i,j,k;
  int t,u;
  int q,r,s;
  scanf("%d",&X);
  scanf("%d",&N);
  for (i=0;i<N;i++){
    scanf("%d",&p[i]);
  }
  /*for (i=0;i<N;i++){
    for (j=i+1;j<N;j++){
      if(p[i]>p[j]){
        t=p[i];
        p[i]=p[j];
        p[j]=t;
      }
    }
  }*/
  /*for(j=0;;j++){
    for(i=0;i<N;i++){
      u=0;
      if( ((X+j)==p[i]) ) u=1;
      if( (X+j)<p[i] ) {
        q=X+j;
        goto loopend1;
      }
      if(u==1) break;
    }
  }
  loopend1:
  for(j=0;;j++){
    for(i=0;i<N;i++){
      u=0;
      if( ((X-j)==p[i]) ) u=1;
      if( (X-j)<p[i] ){
        r=X-j;
        goto loopend2;
      }
    if(u==1) break;
    }
  }
  loopend2:*/
  q=X;
  r=X;
  for(j=0;;j++){
    u=0;
    for(i=0;i<N;i++){
      if(X+j==p[i]){
        u=1;
        break;
      }
    }
    if(u==0){
      q=X+j;
      break;
    }
  }
  for(j=0;;j++){
    u=0;
    for(i=0;i<N;i++){
      if(X-j==p[i]){
        u=1;
        break;
      }
    }
    if(u==0){
      r=X-j;
      break;
    }
  }
  if( (q-X) < (X-r) ){
    s=q;
  } else if( (q-X) > (X-r) ){
    s=r;
  } else if( (q-X) == (X-r) ){
    s=r;
  }
  printf("%d\n",s);
  return 0;
}