A. Gotta Catch Em' All!

time limit per test
1 second
memory limit per test

256 megabytes

input

standard input

output

standard output

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. .

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".

 //2017.01.18
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; const int inf = 0x3f3f3f3f;
int book[]; int main()
{
string str;
string bul = "Bulbbasaur";
while(cin >> str)
{
memset(book, , sizeof(book));
for(int i = ; i < str.length(); i++)
book[str[i]]++;
book['u']/=;
book['a']/=;
int ans = inf;
for(int i = ; i < bul.length(); i++)
if(ans > book[bul[i]])
ans = book[bul[i]];
cout<<ans<<endl;
} return ;
}

CodeForces757A的更多相关文章

随机推荐

  1. js删除最后一个字符串方法

    JS 删除字符串最后一个字符的几种方法 2010-12-02 08:18:35|  分类: 编程 |举报 |字号 订阅   字符串:string s = "1,2,3,4,5," ...

  2. 照着例子学习 protobuf-lua

    参考文章:cocos2dx使用lua和protobuf 首先得下载protobuf-gen-lua的插件,插件Git地址在此. 下载完之后进入到protoc-gen-lua\plugin这个目录,并在 ...

  3. iOS navigationBar 的isTranslucent属性

    苹果文档: A Boolean value indicating whether the navigation bar is translucent (YES) or not (NO). The de ...

  4. Zbus 笔记

    http://blog.csdn.net/cx308679291/article/details/50113257 Zbus学习笔记 标签: zbus 2015-11-30 15:55 266人阅读  ...

  5. python继承的实例

    class SchoolMember(object):#定义学校 member=0#默认成员为0个 amount=0#默认学费为0元 def __init__(self,name,age,sex):# ...

  6. 360路由器设置网段ip

    路由器设置->高级设置->修改路由器地址

  7. linux下源码编译安装mysql

    1.安装依赖的包: yum install -y gdb cmake ncurses-devel bison bison-devel 2.创建mysql安装目录和数据文件目录 mkdir -p /us ...

  8. MUI开发注意事项

    mui开发注意事项,有需要的朋友可以参考下. mui是一个高性能的HTML5开发框架,从UI到效率,都在极力追求原生体验:这个框架自身有一些规则,刚接触的同学不很熟悉,特总结本文:想了解mui更详细的 ...

  9. cocos2dx3.5 HTC One X 某些UI白屏或使用ClippingNode造成部分手机白屏

    public Cocos2dxGLSurfaceView onCreateView() { Cocos2dxGLSurfaceView glSurfaceView = new Cocos2dxGLSu ...

  10. C#中的逆变和协变

    msdn 解释如下: “协变”是指能够使用与原始指定的派生类型相比,派生程度更大的类型. “逆变”则是指能够使用派生程度更小的类型. 解释的很正确,大致就是这样,不过不够直白. 直白的理解: “协变” ...