#ifndef CLASS_DISJOINT_SET
#define CLASS_DISJOINT_SET

#include <vector>
#include <cstdint>
#include <cstring>
#include <algorithm>

class disjoint_set {
private:
	typedef std::int32_t value_type;
	std::vector<value_type> val;
public:
	explicit disjoint_set() : val() {};
	explicit disjoint_set(std::size_t n) : val(n, -1) {};
	std::size_t size() const { return val.size(); }
	std::size_t size(std::size_t elem) { return std::size_t(-val[root(elem)]); }
	std::size_t root(std::size_t elem) {
		// path halving
		while (val[elem] >= 0 && val[val[elem]] >= 0) {
			val[elem] = val[val[elem]];
			elem = val[elem];
		}
		return std::size_t(val[elem] >= 0 ? val[elem] : elem);
	}
	void link(std::size_t elemx, std::size_t elemy) {
		elemx = root(elemx);
		elemy = root(elemy);
		if (elemx == elemy) return;
		if (val[elemx] > val[elemy]) {
			std::swap(elemx, elemy);
		}
		val[elemx] += val[elemy];
		val[elemy] = elemx;
	}
	bool connected(std::size_t elemx, std::size_t elemy) {
		return root(elemx) == root(elemy);
	}
};

#endif // CLASS_DISJOINT_SET

#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
	// step #1. input
	int N;
	cin >> N;
	vector<long long> X(N), Y(N), T(N);
	for (int i = 0; i < N; i++) {
		cin >> X[i] >> Y[i] >> T[i];
	}

	// step #2. calculate distance
	vector<double> radius(N);
	for (int i = 0; i < N; i++) {
		radius[i] = sqrt(X[i] * X[i] + Y[i] * Y[i]);
	}
	auto sqr = [](long long x) {
		return x * x;
	};
	vector<vector<long long> > d(N, vector<long long>(N));
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			d[i][j] = (T[i] == T[j] ? sqr(X[i] - X[j]) + sqr(Y[i] - Y[j]) : ceil((radius[i] - radius[j]) * (radius[i] - radius[j])));
		}
	}

	// step #3. main calculation
	vector<pair<int, int> > edges;
	for (int i = 0; i < N; i++) {
		for (int j = i + 1; j < N; j++) {
			edges.push_back(make_pair(i, j));
		}
	}
	sort(edges.begin(), edges.end(), [&](const pair<int, int>& p1, const pair<int, int>& p2) { return d[p1.first][p1.second] < d[p2.first][p2.second]; });
	disjoint_set uf(N);
	long long answer = -1;
	for (pair<int, int> i : edges) {
		uf.link(i.first, i.second);
		if (uf.connected(0, N - 1)) {
			answer = d[i.first][i.second];
			break;
		}
	}

	// step #4. output
	cout << answer << endl;

	return 0;
}