First, if all numbers are equal, then there is no answer (since $$$a$$$ is divisible by $$$a$$$, if both arrays are non-empty, then $$$c_1$$$ is a divisor of $$$b_1$$$).

Second, if $$$a$$$ is divisible by $$$b$$$ and they are both natural numbers, then the following equality holds: $$$b \le a$$$ (by definition, if $$$a$$$ is divisible by $$$b$$$, then $$$a$$$ can be represented as $$$k \dot b$$$, where $$$k$$$ is a natural number).

Now we can place all instances of the smallest number into $$$b$$$, and all other numbers into $$$c$$$. It can be seen that such a construction always gives a valid answer.

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
 
void solve() {
	int n = 0; cin >> n; 
	vector<int> inp; inp.assign(n, 0);
	for (auto& x : inp) cin >> x;
	sort(inp.begin(), inp.end());
	if (inp.back() == inp[0]) {
		cout << "-1\n";
		return;
	}
	else {
		int it = 0;
		while (inp[it] == inp[0]) it++;
		cout << it << " " << n - it << "\n";
		for (int j = 0; j < it; ++j) cout << inp[j] << " ";
		for (int j = it; j < n; ++j) cout << inp[j] << " ";
	}
}
 
int main() {
	int t = 0; cin >> t;
	while (t--) solve();
	return 0;
}