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>
#include<stdlib.h>
int main(void){
int x,n;
  scanf("%d%d",&x,&n);
  int p[n];
  for(int i=0;i<n;i++){
	scanf("%d",&p[i]);
  }
  if(n==0)printf("%d",x);
  else{
    int list[103];
  for(int i=0;i<103;i++){
	list[i]=i-1;
  }
    for(int i=0;i<n;i++){
	list[p[i]+1]=-3;
    }
    int min=200,minidx[100];
    for(int i=0;i<103;i++){
      if(list[i]==-3)continue;
	if(abs(list[i]-x)<min){
      min=abs(list[i]-x);
    }
    }
    for(int i=0;i<103;i++){
	if(list[i]==-3)continue;
      if(abs(list[i]-x)==min){printf("%d",list[i]);
                              return 0;
                             }
    }
  }
    return 0;
  }