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 <math.h>
#include <string.h>

int main() {
  int x,y,z;
  int p[100];
  int i;
  int a,b;
  a = 101;
  b = -1;
  scanf("%d %d", &x,&y);

  for(i = 0; i< 100; i++) p[i] = i+1;
  for(i = 0; i < y; i++) {
    if(i == y-1) {
      scanf("%d", &z);
      p[z-1] = 0;
    } else {
      scanf("%d ", &z);
      p[z-1] = 0;
    }
  }
  if(y == 100) return 0;
  //for(i = 0; i< 100; i++) printf("%d ", p[i]);
  int flag = 0;
  for(i=x-1; i>=0; i--) {
    if(p[i] != 0) {
      flag = 1;
      break;
    }
  }
  if(flag == 1) a = p[i];
  flag = 0;
  for(i=x-1; i < x+abs(x-a); i++) {
    if(i == 100) break;
    if(p[i] != 0) {
      flag = 1;
      break;
    }
  }
  if(flag == 1) b = p[i];
  if(b == -1 && a != 101) {
    printf("%d\n",a);
  } else if(a == 101 && b != -1) {
    printf("%d\n",b);
  } else if(abs(x-a) == abs(x-b)) {
    printf("%d\n",a);
  } else if(abs(x-a) > abs(x-b)) {
    printf("%d\n",b);
  } else if(abs(x-a) < abs(x-b)) {
    printf("%d\n",a);
  }
  return 0;
}