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<string.h>
#include<stdlib.h>

int main(void){
    int X, N, i, j, k, min=100, min_index=0, flag=0;
    scanf("%d%d", &X, &N);
    int p[N+1], b[101];
    for(i=1; i<=N; i++){
        scanf("%d", &p[i]);
        b[i]=0;
    }
    for(i=1; i<=N; i++){
        b[p[i]]=1;
    }
    for(i=1; i<=100; i++){
        if(N==0){
            printf("%d", X);
            flag=1;
            break;
        }else if(abs(X-i) < min && b[i]==0){
            min = abs(X-i);
            min_index=i;
        }
    }

    if(flag==0){
        printf("%d", min_index);
    }

    return 0;
}