## solution

$N$の約数の個数$d(N)$は最大でも$1000$のオーダーである。 $x, y$についてこれらから総当たりして間に合う。

## implementation

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < int(n); ++ (i))
using ll = long long;
using namespace std;

vector<bool> sieve_of_eratosthenes(int n) {
vector<bool> is_prime(n + 1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= n; ++ i)
if (is_prime[i])
for (int k = 2 * i; k <= n; k += i)
is_prime[k] = false;
return is_prime;
}
vector<int> list_primes(int n) {
auto is_prime = sieve_of_eratosthenes(n);
vector<int> primes;
for (int i = 2; i <= n; ++ i)
if (is_prime[i])
primes.push_back(i);
return primes;
}
map<ll, int> prime_factorize(ll n, vector<int> const & primes) {
map<ll, int> result;
for (int p : primes) {
if (n < p *(ll) p) break;
while (n % p == 0) {
result[p] += 1;
n /= p;
}
}
if (n != 1) result[n] += 1;
return result;
}
vector<ll> list_factors(ll n, vector<int> const & primes) {
vector<ll> result;
result.push_back(1);
for (auto it : prime_factorize(n, primes)) {
ll p; int k; tie(p, k) = it;
int size = result.size();
REP (y, k) {
REP (x, size) {
result.push_back(result[y * size + x] * p);
}
}
}
return result;
}
vector<int> primes = list_primes(1e5);

ll solve(ll n, ll a, ll b, ll c) {
ll cnt = 0;
vector<ll> xs = list_factors(n, primes);
for (ll x : xs) if (x <= a) {
for (ll y : xs) if (y <= b) {
if (n % (x * y) == 0) {
ll z = n / (x * y);
if (z <= c) {
cnt += 1;
}
}
}
}
return cnt;
}

int main() {
int t; cin >> t;
while (t --) {
ll n, a, b, c; cin >> n >> a >> b >> c;
ll result = solve(n, a, b, c);
cout << result << endl;
}
return 0;
}