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: Accepted)

import kotlin.math.abs

fun main(args: Array<String>) {
    val (X, N) = readLine()!!.split(" ").map{it.toInt()}
    if(N < 1) {
      println(X)
      return
    }
    val numbers = readLine()!!.split(" ").map { it.toInt() }.sorted()
  
    val lastNum = numbers.last() + 1
    fun absCalc(counter: Int, minNum: Int, nums: List<Int> = numbers, target: Int = X): Int {
        val newMinNum = if (!nums.contains(counter) && abs(target - counter) <= abs(target - minNum)) {
            counter
        } else minNum

        return if (numbers.first() - 1 > counter) newMinNum else absCalc(counter - 1, newMinNum)
    }
    println(absCalc(lastNum, lastNum))
}