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: Accepted)


fun main(args: Array<String>) {
    val x = getInt()
    val l = lcm(x, 360)
    println(l / x)

}

fun gcd(a: Int, b: Int): Int {
    return if (a % b == 0) b else gcd(b, a % b)
}

fun lcm(a: Int, b: Int): Int {
    return a * b / gcd(a, b)
}

fun getInt() = readLine()!!.toInt()