#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;

ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}

int solve(){
	long long d, ax, ay;
	cin >> d >> ax >> ay;
	long long g = gcd(ax, ay);
	long long k_max = 0;
	long long k_min = 0;
	auto get_max = [&](long long dx, long long dy){
		long long ok = 0, ng = 1001001001;
		while(ng - ok > 1){
			long long mid = (ok + ng) / 2;
			long long bx = ax + dx * mid;
			long long by = ay + dy * mid;
			if(0 <= bx && bx <= d && 0 <= by && by <= d) ok = mid;
			else ng = mid;
		}
		return ok;
	};
	k_max = get_max(-ay / g, ax / g);
	k_min = -get_max(ay / g, -ax / g);
	long long ans = max(k_max, -k_min) * g * ((ax / g) * (ax / g) + (ay / g) * (ay / g));
	cout << ans << "\n";
	return 0;
}

int main(){
	int t;
	cin >> t;
	rep(_, t) solve();
	return 0;
}