Sam loves Lily very much that he shows his love to her through all kinds of ways. This morning, Lily received an e-mail from Sam. Lily knows that Sam hided “good morning” in this mail. Lily tried several ways to resort the letters (including the space ' ') so that more “good morning"s could be found. The number of “good morning” appeared in a specified string equals the number of positions from which Lily could see a consecutive string “good morning”.

With so many letters, Lily is about to be dizzy. She asks you to tell her what is the maximum number of “good morning"s appear in this mail after rearranged in some way.

For every test case, you should output Case #k: first, where $k$ indicates the case number and starts at $1$. Then output an integer indicating the answer to this test case.

2
gninrom doog
ggoooodd mmoorrnniinngg

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=202501;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
//*************************************************************************************int a[30];
char s[maxn];
int main()
{//この戦い终わった、故郷に帰って结婚するだん！！！ ??\cout<<"please hack me"<<endl;int t=read();for(int cas=1;cas<=t;cas++){memset(a,0,sizeof(a));gets(s);for(int i=0;i<strlen(s);i++){if(s[i]<='z'&&s[i]>='a')a[s[i]-'a']++;}int num=inf;num=min(num,a['g'-'a']-1);num=min(num,a['o'-'a']/3);num=min(num,a['d'-'a']);num=min(num,a['m'-'a']);num=min(num,a['r'-'a']);num=min(num,a['n'-'a']/2);num=min(num,a['i'-'a']);printf("Case #%d: %d\n",cas,num);}
}