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

int main(void)
{

  int x, n, p[1000], i, y, j;
  int flag = 0;

  scanf("%d%d", &x, &n);
  for (i = 0; i < n; i++) scanf("%d", &p[i]);

  if (n == 0) printf("%d\n", x);

  /*for (j = 1; j <= n; j++){
    y = x + j;
    for (i = 0; i < n; i++){
      if (y == p[i]) flag = 0;
    }
    y = x - j;
    for (i = 0; i < n; i++){
      if (y == p[i]) flag = 0;
    }
  }*/

  while(1){
    int j = 1;
    y = x + j;
    for (i = 0; i < n; i++){
      if (y == p[i]) flag = 1;
    }
    y = x - j;
    for (i = 0; i < n; i++){
      if (y == p[i]) flag = 1;
    }
    j++;
    if(flag == 0) break;
  }

  printf("%d\n", y);

  return 0;

}