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)

fun main() {
    val inLine = readLine()!!.split(" ").map(String::toInt)
    val x = inLine[0]
    val n = inLine[1]
    var pList = mutableListOf<Int>()
    var result = 0
    if (n == 0){
        result = x
    }else if(!pList.contains(x)){
        result = x
    }else{
        pList = readLine()!!.split(" ").map(String::toInt).toMutableList()

        val lxList = mutableListOf<Int>()
        val bxList = mutableListOf<Int>()
        var minP1 = Integer.MAX_VALUE
        var minP2 = Integer.MAX_VALUE

        pList.forEach{
            if(it > x){
                bxList.add(it)
            }else if(it < x){
                lxList.add(it)
            }
        }
        bxList.add(x)
        lxList.add(x)


        var baseX = x

        if(bxList.isEmpty()){
            minP1 = x + 1
        }else{
            run loop@{
                while(!bxList.isEmpty()){
                    if(!bxList.contains(baseX)){
                        minP1 = baseX
                        return@loop
                    }else{
                        bxList.remove(baseX)
                    }
                    baseX++
                }
            }
        }

        baseX = x
        if(lxList.isEmpty()){
            minP2 = x - 1
        }else{
            run loop@{
                while(!lxList.isEmpty()){
                    if(!lxList.contains(baseX)){
                        minP2 = baseX
                        return@loop
                    }else{
                        lxList.remove(baseX)
                    }
                    baseX--
                }
            }
        }


        result = if(minP1 < minP2) minP1 else minP2
    }
    println(result)
}