D

Suppose there are M people, including you, playing a special card game. At the beginning, each player receives N cards. The pip of a card is a positive integer which is at most N*M. And there are no two cards with the same pip. During a round, each player chooses one card to compare with others. The player whose card with the biggest pip wins the round, and then the next round begins. After N rounds, when all the cards of each player have been chosen, the player who has won the most rounds is the winner of the game.

Given your cards received at the beginning, write a program to tell the maximal number of rounds that you may at least win during the whole game.

Input

The input consists of several test cases. The first line of each case contains two integers m (2�?20) and n (1�?50), representing the number of players and the number of cards each player receives at the beginning of the game, respectively. This followed by a line with n positive integers, representing the pips of cards you received at the beginning. Then a blank line follows to separate the cases.

The input is terminated by a line with two zeros.

Output

For each test case, output a line consisting of the test case number followed by the number of rounds you will at least win during the game.

Sample Input

2 5
1 7 2 10 9

6 11
62 63 54 66 65 61 57 56 50 53 48

0 0

Sample Output

Case 1: 2
Case 2: 4

这题只要比较其他人中有多少可以比他大的，他如果是最大的就不管他，跳过就好了，田忌赛马哈哈；

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define cl clear()
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
const double pi=acos(-1.0);
typedef __int64 ll;
typedef long double ld;
const ll MOD=1e9+7;
using namespace std;
bool com(int a,int b)
{
	return  a>b;
}
int a[1005],d[1005];
int main()
{
	vector<int> v;
	int cas=1,m,n;
	while(1)
	{
		v.cl;
		sf("%d%d",&m,&n);
		if(m==0)
		return 0;
		for(int i=0;i<n;i++)
		sf("%d",&a[i]);
		sort(a,a+n,com);
		int i,j;
		for( i=n*m,j=0;i>0;i--)
		{
			if(i!=a[j])
			v.pb(i);
			else
			j++;
		}
		int ans=0;
		for(i=0,j=0;i<n;)
		{
			if(a[i]>v[j])
			i++;
			else
			{
				ans++;
				i++;
				j++;
			}
		}
		pf("Case %d: %d
",cas++,n-ans);
	}
}