#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#include <atcoder/math>
using namespace atcoder;

// M 以下の正の整数のうち、A で割ったあまりが C となるものの個数
ll f(ll M, ll A, ll C)
{
  if (C != 0)
    M += A - C;
  return max(0LL, M / A);
}

ll solve(ll M, ll A, ll B, ll K)
{
  ll g = gcd(A, B);
  if (K % g != 0)
    return 0;
  M /= g, A /= g, B /= g, K /= g;
  
  if (K > A)
    return 0;
  else if (K == A)
  {
    ll cnt_A = f(M, A, 0), cnt_B = f(M, B, 0), cnt_AB = 0;
    ll N = cnt_A + cnt_B - cnt_AB;
    ll cnt_notA = cnt_B - cnt_AB;
    ll X_N = max(M / A * A, M / B * B);
    if (X_N % A == 0)
      return (N - 1) - (2 * cnt_notA);
    else
      return (N - 1) - (2 * cnt_notA - 1);
  }
  else
  {
    ll Y = crt({0, -K}, {A, B}).first;
    ll Z = crt({-K, 0}, {A, B}).first;
    return f(M - K, A * B, Y) + f(M - K, A * B, Z);
  }
}

int main()
{
  int T;
  cin >> T;
  while (T--)
  {
    ll M, A, B, K;
    cin >> M >> A >> B >> K;
    cout << solve(M, A, B, K) << endl;
  }
}