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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import kotlin.math.abs

fun main() {
    val (x, n) = readLine()!!.split(" ").map(String::toInt)
    if (n == 0) {
        println(x)
        return
    }
    val nums = readLine()!!.split(" ").map(String::toInt).sorted()
    var near: Int? = null
    (nums.first()..nums.last()).forEach {
        if (it !in nums) {
            val currentNear = near ?: run {
                near = it
                return@forEach
            }
            if (abs(x - it) < abs(x - currentNear)) {
                near = it
            }
        }
    }
    println(near)
}