The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.

For
a few months now, Roy has been assessing the security of various banks
and the amount of cash they hold. He wants to make a calculated risk,
and grab as much money as possible.

His mother, Ola, has decided upon a tolerable
probability of getting caught. She feels that he is safe enough if the
banks he robs together give a probability less than this.

 

The first line of input gives T, the number of cases. For each
scenario, the first line of input gives a floating point number P, the
probability Roy needs to be below, and an integer N, the number of banks
he has plans for. Then follow N lines, where line j gives an integer Mj
and a floating point number Pj .

Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .

 

For each test case, output a line with the maximum number of millions
he can expect to get while the probability of getting caught is less
than the limit set.

Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all
probabilities are independent as the police have very low funds.

 

 

   /* ***********************************************

   Author        :Mubaixu

   File Name     :HDU2955.cpp

   ************************************************ */

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<queue>

#include<algorithm>

using  namespace std;

const int maxm=;

const int maxn=;

double dp[maxm];

int value[maxn];

double tp[maxn];

int main(){

     int t;

    int n;

    double p;

    scanf("%d",&t);

    while(t--){

         scanf("%lf%d",&p,&n);

        int sum=;

        for(int i=;i<=n;i++){

             scanf("%d%lf",&value[i],&tp[i]);

            tp[i]=-tp[i];

            sum+=value[i];

        }

         memset(dp,,sizeof(dp));

        dp[]=;

        for(int i=;i<=n;i++){

           for(int j=sum;j>=value[i];j--){

               dp[j]=max(dp[j],dp[j-value[i]]*tp[i]);

           }

        }

          for(int i=sum;i>=;i--){

             if(dp[i]>(-p)){

                 printf("%d\n",i);

                 break;

             }

          }

    }

    return ;

}