B - Vicious Keyboard

B - Vicious Keyboard 

Tonio has a keyboard with only two letters, "V" and "K".

One day, he has typed out a string s with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.

Input

The first line will contain a string s consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.

Output

Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.

Example

Input

VK

Output

1

Input

VV

Output

1

Input

V

Output

0

Input

VKKKKKKKKKVVVVVVVVVK

Output

3

Input

KVKV

Output

1
题意：输入一个字符串，输出其中有多少对VK，可以任意更改其中的一个字符串使VK对的数量最多。
题解：（暴力）先不更改计算有多少对VK，然后没次更改一个字符串计算一次，取最大值。
代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;
int main()
{
    char str[105],s;
    int ans[210],i,j,len;
    while(cin>>str){
           memset(ans,0,sizeof(ans));
        len=strlen(str);
        for(i=0;i<len;i++)
            if(str[i]=='V'&&str[i+1]=='K'&&(i+1)<len)
            ans[0]++;
        for(j=1;j<=len;j++)//依次把字符串换成K计算，每次换完后要换回去。
        {
            s=str[j-1];
            str[j-1]='K';
                for(i=0;i<len;i++)
            if(str[i]=='V'&&str[i+1]=='K'&&(i+1)<len)
            ans[j]++;
            str[j-1]=s;
        }
        for(j=len+1;j<=2*len;j++)//依次把字符串换成V计算
        {
            s=str[j-1-len];
            str[j-1-len]='V';
            for(i=0;i<len;i++)
            if(str[i]=='V'&&str[i+1]=='K'&&(i+1)<len)
            ans[j]++;
            str[j-1-len]=s;
        }
        ans[0]=*max_element(ans,ans+2*len+1);
        cout<<ans[0]<<endl;
    }
}