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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
// Your code here!
fscanf(STDIN,"%d %d",$x,$n);
$p=explode(' ',trim(fgets(STDIN)));

// $nが0の場合
if($n == 0){
    echo $x;
    exit;
}

// 検索対象が配列内に存在しない場合
if(!in_array($x, $p)){
    echo $x;
    exit;
}

$i=1;
// マイナス方向で探索
while(true){
    $num=$x-$i;
    var_dump($num);
    if(!in_array($num, $p)){
        break;
    }
    $i+=1;
}

$j=1;
// ＋方向で探索
while(true){
    $num=$x+$j;
    var_dump($num);
    if(!in_array($num, $p)){
        break;
    }
    $j+=1;
}

if($i<=$j){
    echo $x-$i;
}else{
    echo $x+$j;
}
?>