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: Wrong Answer)

#include <stdlib.h>
#include <stdio.h>

int main(){
  int x, n, flag1, flag2;
  scanf("%d", &x);
  scanf("%d", &n);
  int p[n];
  for(int i=0; i < n;i++){
    scanf("%d", &p[i]);
  }
  if(n==0){
    printf("%d", x);
  }else{
  for(int j=1;j<256;j++){
    flag1=0;
    flag2=0;
    for(int k = 0;k<n;k++){
      if(p[k] == x+j){
        flag1=1;
      }
      if(p[k] == x-j){
        flag2=1;
      }
      if(flag1==0 || flag2==0){
        break;
      }
    }
    if(flag1==1&&flag2==0){
      printf("%d", x-j);
      break;
    }else if(flag1==0&&flag2==1){
      printf("%d", x+j);
      break;
    }else if(flag1==1&&flag2==1){
      continue;
    }else{
      printf("%d", x-j);
      break;
    }
  }
  }
  return 0;
}