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() {
    agc046a()
}

fun agc046a() {
    val x = readLine()!!.toInt()
    var t = x
    var count = 1
    while (t != 0) {
        t += x
        t %= 360
        count++
    }
    println(count)
}