Problem:
Score : 200 points

Problem Statement

Takahashi is standing on a two-dimensional plane, facing north. Find the minimum positive integer K such that Takahashi will be at the starting position again after he does the following action K times:

Go one meter in the direction he is facing. Then, turn X degrees counter-clockwise.

Constraints

1 \leq X \leq 179

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the number of times Takahashi will do the action before he is at the starting position again.

Sample Input 1

90

Sample Output 1

4

Takahashi's path is a square.

Sample Input 2

1

Sample Output 2

360

Solution in Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main() {
    val exteriorangle = readLine()!!.trim().toInt()
    val interiorangle=180-exteriorangle
    var rest:Int=360%exteriorangle
    var ans:Int = 360/exteriorangle
    var n:Int = 1
     while(rest!=0){

         var sumoftheinteriorangle = n*interiorangle
         rest=sumoftheinteriorangle%180
         n++
         ans=n-1
     }
    println(ans)
}