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 (Clang 10.0.0) (judge verdict: Accepted)

#include <stdio.h>

int main()
{

    int a;
    int b;

    
    int n[101];
    int deta[101];

    scanf("%d", &a);
    scanf("%d", &b);

    for (int i = 0; i <= 100; i++)
    {
        deta[i] = 0;//で初期化
    }

    for (int i = 0; i < b; i++)
    {
        scanf("%d", &n[i]);
        deta[n[i]] = 1;
    }

    for (int i = 0; i <= 100; i++)
    {
        if(a-i>=0){//0よりでかい
        if(deta[a-i]==0){
            printf("%d", a - i);
            break;
        }
        }
        if(a+i==101){
            printf("%d", 101);
            break;
        }
        if (a + i <= 100)
        { //100より
            if (deta[a + i] == 0)
            {
                printf("%d", a + i);
                break;
            }
        }
    }

    printf("\n");
}