/*
	author: TangQiao , Wind @ Beijing Normal University

	problem name: Bullshit Bingo
	
	source :  PKU Online Judge
	
	problem type: 字符串处理题
	
	problem description: 以BULLSHIT为界,找出每组中不重复的字符串的个数,最后输出
	各组和 / 组数.要求约分.
	
	problem solution: 模拟找就行了.
	

	faults: 当前此题的代码只能在G++的编译器下提交通过.具体原因不明.
	
	date : 2005.7.10 北大个人赛
	
*/

#include <stdio.h>
#include <string.h>


char ss[910][50];
int match[50000000];
int n_match=0;

void output(int a,int b)
{
	int i;
	int min;
	min=a>b?b:a;
	
	if (min==0)
		printf("%d / %d\n",a,b);
	else
	{	
		for (i=min;i>=2;i--)
			if (a%i==0 && b%i==0)
			{
				a/=i;
				b/=i;
				i++;
			}
		printf("%d / %d\n",a,b);
	}
}

main()
{
	int i,j;
	char in[30], ch;
	int head;
	int num;
	int over;
	char bb[30];

//	freopen("bingo.in", "r", stdin);

	num=0;
	head=0;
	over=0;	
	//strcpy(in,kong);
	memset(in,0,sizeof(in));
	memset(bb,0,sizeof(bb));
	strcpy(bb,"BULLSHIT");
	while (scanf("%c", &ch)!=EOF)

	{
//		if (ch>='a' && ch<='z') 
//			ch+='A'-'a';
		if ((ch>='A' && ch<='Z' ) || (ch>='a' && ch<='z'))
		{
			in[head++]=ch;
		}
		else
		{ 
			if (head!=0)
			{
				in[head]=0;
				num++;
				strcpy(ss[num],in);
				if (strcmp(in,bb)==0) over=1;

		//		strcpy(in,kong);
				memset(in,0,sizeof(in));
				head=0;
			}
		}

		if (over)
		{
			num--;
			over=0;
			n_match++;
			match[n_match]=num;

			for (i=1;i<=num;i++)
			{
				int len;
				len=strlen(ss[i]);
				for (j=0; j < len; j++)
					if (ss[i][j]>='a' && ss[i][j]<='z') 
						ss[i][j]+=('A'-'a');
			}
			for (i=1;i<num;i++)
			{
				for (j=i+1;j<=num;j++)
					if (strcmp(ss[i],ss[j])==0)
					{
						match[n_match]--;
						break;
					}
			}
			num=0;
			head=0;
		}
	} // end of while 

	if (head!=0)
	{
		in[head]=0;
		num++;
		strcpy(ss[num],in);
		if (strcmp(in,bb)==0) over=1;
		
	//	strcpy(in,kong);
		memset(in,0,sizeof(in));
		head=0;
	}
	if (over)
	{
		num--;
		over=0;
		n_match++;
		match[n_match]=num;
		for (i=1;i<num;i++)
		{
			for (j=i+1;j<=num;j++)
				if (strcmp(ss[i],ss[j])==0)
				{
					match[n_match]--;
					break;
				}
		}
		num=0;
		head=0;
	}

	for (i=1,j=0;i<=n_match;i++)
	{
		j+=match[i];
//		printf("num:%d tot=%d\n", i, match[i]);
	}
	output(j,n_match);
	

	return 0;
}