#include <stdio.h>
#include <vector>
#include <algorithm>
#include <queue>
#define maxn 10002
typedef long long ll;
using namespace std;
struct round
{
  ll time;
  int sign;
  int idx;
};
int n;
int r[maxn], d[maxn], w[maxn];
round pro[maxn * 2];
ll remain[maxn], deadline[maxn];
bool operator < (const round &A, const round &B)
{
  return A.time < B.time;
}
struct deadline_cmp
{
  bool operator () (const int &A, const int &B) {
    return deadline[A] > deadline[B];
  }
};
bool solve(ll speed)
{
  vector<ll> xs;
  for(int i=1; i<=n; ++i)
  {
    remain[i] = w[i];
    deadline[i] = speed * d[i];
    // 시작 시간과 데드라인을 모음 
    pro[i*2 - 1].time = speed * r[i];
    pro[i*2 - 0].time = speed * d[i];
    pro[i*2 - 1].sign = 1;
    pro[i*2 - 0].sign = -1;
    pro[i*2 - 1].idx = pro[i*2].idx = i;
    xs.push_back( speed * r[i]);
    xs.push_back( speed * d[i]);
  }
  // 모든 시각을 모아서 소트하고 중복제거
  sort(pro + 1, pro + 2*n + 1);
  sort(xs.begin(), xs.end());  
  xs.erase(unique(xs.begin(), xs.end()), xs.end());
  // deadline 힙
  priority_queue < int, vector<int>, deadline_cmp> heap;
  for(int ptr=1, i=0; i < xs.size(); ++i)
  {
    while(ptr <= 2*n && pro[ptr].time == xs[i])
    {
      // 새롭게 작업이 추가되었으므로 HEAP으로
      if(pro[ptr].sign == 1)
        heap.push(pro[ptr++].idx);
      // 데드라인을 만났는데 아직 안끝났으면 불가능한 경우!
      else if (remain[ pro[ptr++].idx ])
        return false;
    }
    if(i == xs.size() - 1) break;
    ll times = xs[i + 1] - xs[i];
    while(times > 0 && !heap.empty())
    {
      // 힙에서 가장 deadline이 임박한 작업을 꺼내서 process
      int now = heap.top(); heap.pop();
      if(remain[now] <= times)
      {
        times -= remain[now];
        remain[now] = 0;
      }
      else
      {
        remain[now] -= times;
        times = 0;
        heap.push(now);
      }
    }
  }
  return true;
}
int main()
{
  int T;
  for( scanf("%d", &T); T > 0; -- T )
  {
    scanf("%d", &n);
    for(int i=1; i<=n; ++i)
      scanf("%d %d %d", r+i, d+i, w+i);
    // 이분 검색 부분
    ll left = 0, right = 2147483647;
    ll mid, answer;
    while(left <= right)
    {
      mid = (left + right) / 2;
      if( solve(mid) )
      {
        answer = mid;
        right = mid - 1;
      }
      else left = mid + 1;
    }
    printf("%d\n", (int)answer);
  }
  return 0;
}