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<string.h>
#include<stdbool.h>
#include<math.h>
#include<limits.h>
int main(){
    int X,N;
    (void)scanf("%d %d",&X,&N);
    int p[102];
    int f[101];
    memset(f,0,sizeof(f));
    for(int i=0;i<N;i++){
        (void)scanf("%d",p+i);
        f[p[i]] = 1;
    }
    int def = 0;
    for(;;){
        if(X-def<1){(void)printf("%d",X-def); return 0;}
        else if(X+def>100){(void)printf("%d",X+def); return 0;}
        else if(f[X-def]==0){(void)printf("%d",X-def);return 0;}
        else if(f[X+def]==0){(void)printf("%d",X+def);return 0;}
        else def++;
    }
}