/*
  Solution to problem: 181 - Scuba diver
  (From the Sphere Online Judge -  http://www.spoj.pl)

  Author: Andres Mejia-Posada
  http://blogaritmo.factorcomun.org
 */
#include <iostream>

using namespace std;

int main(){
  int C;
  cin >> C;
  while (C--){
    int O, N, Y;
    cin >> O >> N >> Y;
    int w[Y+1], oxy[Y+1], nitro[Y+1];
    for (int i=1; i<=Y; ++i){
      cin >> oxy[i] >> nitro[i] >> w[i];
    }

    //dp[i][o][n] = minimum weight needed to have o-oxygen and n-nitrogen (or more) using the first i cylinders.
    unsigned int dp[Y+1][O+1][N+1];
    for (int i=0; i<=Y; ++i){
      for (int o=0; o<=O; ++o){
        for (int n=0; n<=N; ++n){
          dp[i][o][n] = INT_MAX;
        }
      }
    }

    dp[1][0][0] = 0;
    for (int o=1; o<=oxy[1]; ++o){
      for (int n=1; n<=nitro[1]; ++n){
        dp[1][o][n] = w[1];
      }
    }

    for (int i=2; i<=Y; ++i){
      for (int o=0; o<=O; ++o){
        for (int n=0; n<=N; ++n){
          dp[i][o][n] = min(dp[i-1][o][n], dp[i-1][max(0, o-oxy[i])][max(0, n-nitro[i])] + w[i]);
        }
      }
    }

    cout << dp[Y][O][N] << endl;

  }
  return 0;
}