codeforces 757 A. Gotta Catch Em' All!

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本文介绍了一个有趣的算法问题，即如何使用报纸上的字母拼出尽可能多的“Bulbasaur”这个词。通过统计字符串中特定字母的数量并考虑大小写敏感性，文章提供了一种高效的解决方案。

Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.

Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.

Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?

Note: uppercase and lowercase letters are considered different.

Input

Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.

The string s contains lowercase and uppercase English letters, i.e. $\text{[math]}$ .

Output

Output a single integer, the answer to the problem.

Examples

input
Bulbbasaur

output
1

input
F

output
0

input
aBddulbasaurrgndgbualdBdsagaurrgndbb

output
2

Note

In the first case, you could pick: Bulbbasaur.

In the second case, there is no way to pick even a single Bulbasaur.

In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".

题意：给你一个字符串，问用这些字母可以拼出多少个Bulbasaur...

字母数目数错真是爽...

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define N 200020
using namespace std;
map<char,int>M;
char t[7]={'B','u','l','b','a','s','r'};
int num[7]={1,2,1,1,2,1,1};
char ch[N];
int main()
{
	scanf("%s",ch);int n=strlen(ch);
	for(int i=0;i<n;i++)M[ch[i]]++;
	int ans=1<<30;
	for(int i=0;i<7;i++)ans=min(ans,M[t[i]]/num[i]);
	printf("%d\n",ans);
}