#include<stdio.h>
int main()
{
	int n,  b; int a[100000]; int c = 0; int d = 0;
	scanf("%d", &n);
	for (int i = 2; i <10000; i++)
	{
		for (int j = 2; j <i; j++)
		{
			b = i % j;
			if (b == 0)
				break;
		}
		if (b != 0)
		{
			a[c] = i;    //存质数
			c++;         
		}
	}
	for (int r = 0; r < 10000; r++)
	{
		for (int t = 0; t <10000; t++)
		{
			d = a[r] * a[t];
			if (n == d)  //满足条件输出较大值
			{
					printf("%d", a[t]);
					return 0;
			}
			d = 0;//重置d的值
		}
	}
	return 0;
}