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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#define gc getchar_unlocked
#define pc putchar_unlocked

int ini() {
  int v = 0;
  for (char c = gc(); '0' <= c && c <= '9'; c = gc())
    v = v * 10 + (c - '0');
  return v;
}

void puti(int v) {
  char b[10]; int n = 0;
  while (v)b[n++] = '0' + v % 10, v /= 10;
  if (!n)b[n++] = '0';
  while (n--)pc(b[n]);
}

int main() {
  int x = ini();
  for (int i = 360;; i += 360) {
    if (i % x == 0) {
      puti(i / x);
      return 0;
    }
  }
}