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>
signed main(void)
{
  int X,N,a,dis=100,P[100]={0},min=100,save=100;
  if(scanf("%d %d",&X,&N)>0)
  {
    for(int i=0;i<100;i++){P[i]=i+1<X?X-i-1:i+1-X;}
    while(N--)
    {
      if(scanf("%d",&a)==1){P[a-1]=101;}
    }
    for(int i=0;i<100;i++)
    {
      if(P[i]<101)
      {
        dis=dis<P[i]?dis:P[i];
        if(dis==P[i]){save=i+1<X?X-dis:X+dis;min=min<save?min:save;}
      }
    }
    return printf("%d\n",save),0;
  }
}