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 <malloc.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

//ABC170D
//ABC170C
int main(void) {
  int x, n, k, i, cou=0, tmp;
  scanf("%d %d", &x, &n);
  int p[n];
  for(i=0; i<n; i++){
    scanf(" %d", &p[i]);
  }

  if(n==0){
    printf("%d", x);
    return 0;
  }

  for(i=0; i<n; i++){
    if(p[i]==x) break;
    else if(i==n-1){
      printf("%d", x);
      return 0;
    }
  }

  for(k=1; k<110; k++){
    for(i=0; i<n; i++){
      if(p[i]==x-k) break;
      else if(i==n-1){
        printf("%d", x-k);
        return 0;
      }
    }
    for(i=0; i<n; i++){
      if(p[i]==x+k) break;
      else if(i==n-1){
        printf("%d", x+k);
        return 0;
      }
    }
  }
}