#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <string>
#include <queue>
#include <map>
#include <bitset>
#include <set>
#include <stack>
#include <numeric>
#include <unordered_map>
#include <random>

using namespace std;

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vs = vector<string>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using vpdd = vector<pdd>;
const int inf = (1 << 30) - 1;
const ll INF = 1LL << 60;
//const int MOD = 1000000007;
const int MOD = 998244353;

vector<pii> prime_factorize(ll N) {
	vector<pii> res;
	for (ll a = 2; a * a <= N; ++a) {
		if (N % a != 0) continue;
		ll ex = 0;

		while (N % a == 0) {
			++ex;
			N /= a;
		}
		res.push_back({ a, ex });
	}
	if (N != 1) res.push_back({ N, 1 });
	return res;
}

template<class T1, class T2>
ostream& operator <<(ostream& s, pair<T1, T2> P) {
	return s << P.first << " " << P.second;
}

int main() {
	int q;
	cin >> q;
	int mod = 100000009;

	while (q--) {
		int seed, n, k, b;
		cin >> seed >> n >> k >> b;
		vl x(n + 1);
		x[0] = seed;
		for (int i = 0; i < n; i++) {
			x[i + 1] = 1 + (x[i] * x[i] + 12345 * x[i]) % mod;
		}
		// Bの素因数が一組揃うとB進数で0になる
		// 例として、B=12(2^2*3^1)の時、Tの素因数について
		// min(2の数/2, 3の数) が答えになるので、
		// それぞれの素因数について、Xの小さい方からK個分
		// 選んだ時の最小値を探す
		vpii pb = prime_factorize(b);

		int ans = inf;
		for (auto [p, num] : pb) {
			vi cnt(n + 1, 0);
			for (int i = 0; i <= n; i++) {
				int tmp = x[i];
				while (x[i] % p == 0) {
					x[i] /= p;
					cnt[i]++;
				}
			}

			sort(cnt.begin(), cnt.end());
			int sum = 0;
			for (int i = 0; i < k; i++) {
				sum += cnt[i];
			}
			ans = min(ans, (int)(sum / num));
		}
		cout << ans << endl;

	}

	return 0;
}