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 java.util.*
import kotlin.math.*
import java.io.BufferedReader
import java.io.InputStreamReader


private var fs = BufferedReader(InputStreamReader(System.`in`))
private fun readLn() = fs.readLine()!! // string line
private fun readStrings() = readLn().split(" ") // list of strings
private fun readLong() = readLn().toLong()
private fun readLongs() = readStrings().map { it.toLong() } // list of ints
private fun readInt() = readLn().toInt() // single int
private fun readInts() = readStrings().map { it.toInt() } // list of ints

fun main() {
    solveAtc()
}
fun solveAtc() {
    var (x,n) = readInts()
    if(n == 0){
        println(x)
        return
    }
    var a = readInts()
    var best= Int.MAX_VALUE
    var ans= Int.MAX_VALUE
    for(i in -50..110){
        if (!a.contains(i) && best > abs(x - i)) {
            ans = i
            best = abs(x-i)
        }
        else if(best == abs(x-i)) ans = min(ans,i)
    }
    println(ans)
}