As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermione and Ron have decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives an integer M_j (0 < M_j ≤ 100) and a real number P_j . Bank j contains M_jmillions, and the probability of getting caught from robbing it is P_j. A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output

For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

Note

For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (1 - 0.02) * .03 = 0.0494which is greater than the given probability (0.04). That‘s why he has only option, just to rob rank 2.

题解：给你一个总概率p，再给你N个银行，里面有钱vi，被抓到的概率是pi，要使被抓住的概率小于p，也就是没被抓到的概率大于1-p就好了，总钱数当作背包的容量，概率当作背包的值；

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
typedef long long LL;
#define mem(x,y) memset(x,y,sizeof(x))
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define P_ printf(" ")
#define T_T while(T--)
const int MAXN=110;
struct Node{
	double p;
	int w;
	friend bool operator < (Node a,Node b){
		if(a.p<b.p)return true;
		else return false;
	}
};
double bag[10010];
Node dt[MAXN];
int main(){
	int T,kase=0;
	SI(T);
	T_T{
		double p;
		int N;
		scanf("%lf%d",&p,&N);
		int sum=0;
		for(int i=0;i<N;i++)scanf("%d%lf",&dt[i].w,&dt[i].p),sum+=dt[i].w;
		sort(dt,dt+N);
		mem(bag,0);bag[0]=1;
		for(int i=0;i<N;i++){
			for(int j=sum;j>=dt[i].w;j--)
				bag[j]=max(bag[j],bag[j-dt[i].w]*(1-dt[i].p));
		}
		for(int i=sum;i>=0;i--){
			if(bag[i]>1-p){
				printf("Case %d: %d\n",++kase,i);break;
			}
		}
	}
	return 0;
}