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>

int main(){
int X,N;
int z=0;
int flag[101];
  scanf("%d %d",&X,&N);
  int *p=(int*)malloc(sizeof(int)*N);
  for(int i=0;i<101;i++)flag[i]=1;
  for(int i=0;i<N;i++){
  	scanf("%d",&p[i]);
  }
  for(int i=0;i<N;i++)flag[p[i]]=0;
  for(z=0;z<100;z++){
   if(X-z>0 &&flag[X-z]==1){
         printf("%d\n",X-z);
         break;
   }
   else if(X+z<101 && flag[X+z]==1){
       printf("%d\n",X+z);
       break;
   }
    if(X-z<=0)printf("%d\n",X-z);
    else if(X+z>=101)printf("%d\n",X+z);
  }
  return 0;
}