学步园

//
//完全背包问题
//DP[v]为重量为v的钱罐最少有多少钱 
//DP[v] = min{DP[v],DP[v-W[i]] + P[i]}
//其实就是 DP[i,v] = min{DP[i-1,v],DP[i,v-W[i]] + P[i]}
//也可以用贪心来做
#include <iostream>
#include <cstring>
#define INF 9999999
using namespace std;

int T;
int E,F;
int N;
long P[50001],W[10001];
long DP[10001];

int main()
{
	cin>>T;
	while(T--)
	{
		cin>>E>>F;
		cin>>N;
		for(int i = 1; i <= N; i++)
		{
			cin>>P[i]>>W[i];
		}
		memset(DP,INF,sizeof(DP));
		DP[0] = 0;
		for(int i = 1; i <= N; i++)
		{
			for(long j = W[i]; j <= F-E; j++)
			{
				DP[j] = min(DP[j],DP[j-W[i]] + P[i]);
			}
		}
		if(DP[F-E] < INF)
			cout<<"The minimum amount of money in the piggy-bank is "<<DP[F-E]<<"."<<endl;
		else
			cout<<"This is impossible."<<endl;

	}
	return 0;
}