Codecraft-17 and Codeforces Round #391 (Div. 1 + Div. 2, combined)A+B

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”.
题意：问字符串能能组成多少个”Bulbasaur“。
题解：简单题。
代码：

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <string.h>#include <algorithm>#include <math.h>#include <map>#define ll long longusing namespace std;string str;int main(){    cin>>str;    int B,l,b,a,s,u,r;    b=B=l=a=s=u=r=0;    int len=str.length();    for(int i=0;i<len;i++)    {        if(str[i]=='b')            b++;        if(str[i]=='B')            B++;        if(str[i]=='u')            u++;        if(str[i]=='l')            l++;        if(str[i]=='a')            a++;        if(str[i]=='s')            s++;        if(str[i]=='r')        r++;    }    int mx=10000000;    mx=min(mx,min(b,B));    mx=min(mx,min(u/2,l));    mx=min(mx,min(a/2,s));    mx=min(mx,r);    cout<<mx<<endl;}

B. Bash’s Big Day

time limit per test:2 seconds

memory limit per test:512 megabytes

input:standard input

output:standard output

Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu’s Lab. Since Bash is Professor Zulu’s favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.

But Zulu warns him that a group of k > 1 Pokemon with strengths {s1, s2, s3, …, sk} tend to fight among each other if gcd(s1, s2, s3, …, sk) = 1 (see notes for gcd definition).

Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?

Note: A Pokemon cannot fight with itself.

Input

The input consists of two lines.

The first line contains an integer n (1 ≤ n ≤ 105), the number of Pokemon in the lab.

The next line contains n space separated integers, where the i-th of them denotes si (1 ≤ si ≤ 105), the strength of the i-th Pokemon.

Output

Print single integer — the maximum number of Pokemons Bash can take.

Examples

Input
3
2 3 4

Output
2

Input
5
2 3 4 6 7

Output
3

Note

gcd (greatest common divisor) of positive integers set {a1, a2, …, an} is the maximum positive integer that divides all the integers {a1, a2, …, an}.

In the first sample, we can take Pokemons with strengths {2, 4} since gcd(2, 4) = 2.

In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with gcd ≠ 1.
题意：求出最长子序列使得gcd（a1。。ax）！=1
题解：先处理出每个因子对答案的贡献。然后o（n）求出最大值即可。
代码：

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <string.h>#include <algorithm>#include <math.h>#include <map>#define ll long longusing namespace std;int a[100010];int vis[100010];int ans[100010];void solve(){    memset(ans,0,sizeof(ans));    memset(a,0,sizeof(a));    for(int i=2;i<=100000;i++)    {        if(a[i]==0)        {            ans[i]+=vis[i];            for(int k=i*2;k<=100010;k+=i)            {                ans[i]+=vis[k];                a[k]=1;            }        }    }}int n,x;int main(){    scanf("%d",&n);    memset(vis,0,sizeof(vis));    for(int i=1;i<=n;i++)    {        scanf("%d",&x);        vis[x]++;    }    solve();    int op=1;    for(int i=1;i<=100000;i++)    {        op=max(op,ans[i]);    }    printf("%d\n",op);}