#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <climits>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
#define REP(i,n) for(int i=0; i<n; i++)
#define FOR(i,st,end) for(int i=st;i<end;i++)
#define db(x) cout << (#x) << " = " << x << endl;
#define mp make_pair
#define pb push_back
typedef long long int ll;

const int MAX = 1024;
const int INF = 999999999;

int N, O, n, t, cs;
int nitro[MAX], oxi[MAX], w[MAX];
int dp[MAX][22][80], visited[MAX][22][80];

int solve(int i, int co, int cn)
{
	if(dp[i][co][cn]!=-1){
		return dp[i][co][cn];
	}
	if(co==0&&cn==0){
		dp[i][co][cn]= 0;
	}
	else if(i==0){//if we considered all cylinders and if co and cn are still not equal to zero
		//then its not possible to satisfy the required condition. Hence return INF in order to eliminate all
		//recursive paths leading to this status
		dp[i][co][cn]= INF;
	 }else{
		//either select the ith cylinder or move on to i-1th cylinder
		//Be careful about the index:
		//ith cylinder oxi[i-1] capacity of oxygen and nitro[i-1] capacity of nitrogen
		dp[i][co][cn]=min(solve(i-1,co,cn),solve(i-1,max(co-oxi[i-1],0),max(cn-nitro[i-1],0))+w[i-1]);
	}
	return dp[i][co][cn];
}

int main()
{
	scanf("%d", &t);
	while(t--){

		scanf("%d %d %d", &O, &N, &n);
		for(int i=0;i<n+1;i++)
			for(int j=0;j<O+1;j++)
				for(int k=0;k<N+1;k++)
					dp[i][j][k]=-1;

		for(int i = 0; i < n; i++)
            scanf("%d %d %d", &oxi[i], &nitro[i], &w[i]);
		//dp[n][O][N] gives the minimum total weight of cylinders required to have O and N capacity of oxygen
		//and nitrogen (capacity of oxygen and nitrogen can exceed)
		printf("%d\n", solve(n, O, N));
	}
	return 0;
}