Problem:
Problem D: Circle and Points

You are given N points in the xy-plane. You have a
circle of radius one and move it on the xy-plane, so as to
enclose as many of the points as possible. Find how many points can be
simultaneously enclosed at the maximum. A point is considered
enclosed by a circle when it is inside or on the circle.

Fig 1. Circle and Points

Input

The input consists of a series of data sets, followed by a single line
only containing a single character '0', which indicates the end of the
input. Each data set begins with a line containing an integer N,
which indicates the number of points in the data set. It is
followed by N lines describing the coordinates of the
points. Each of the N lines has two decimal fractions X
and Y, describing the x- and y-coordinates of a
point, respectively. They are given with five digits after the decimal
point.

You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0
<= Y <= 10.0. No two points are closer than 0.0001. No two
points in a data set are approximately at a distance of 2.0. More
precisely, for any two points in a data set, the distance d
between the two never satisfies 1.9999 <= d <= 2.0001. Finally,
no three points in a data set are simultaneously very close to a
single circle of radius one. More precisely,
let P1, P2, and P3 be any three points in a
data set, and d1, d2, and d3
the distances from an arbitrarily selected point in the xy-plane to each of them
respectively. Then it never simultaneously holds that 0.9999 <=
di <= 1.0001 (i = 1, 2, 3).

Output

For each data set, print a single line containing the maximum number
of points in the data set that can be simultaneously enclosed by a
circle of radius one. No other characters including leading and
trailing spaces should be printed.

Sample Input

3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
0

Output for the Sample Input

2
5
5
11

Solution in C++11 (judge verdict: Wrong Answer)

////////////////////
///   template   ///
////////////////////
#include <cassert>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <numeric>
#include <functional>
#include <vector>
#include <queue>
#include <string>
#include <complex>
#include <stack>
#include <set>
#include <map>
#include <list>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
using namespace std;

//// MACRO ////
#define REP(i,n) for (int i = 0; i < (n); i++)
#define RREP(i,n) for (int i = (n)-1; i >= 0; i--)
#define FOR(i,s,n) for (int i = (s); i < (n); i++)
#define allof(c) c.begin(), c.end()
#define partof(c,i,n) c.begin() + (i), c.begin() + (i) + (n)
#define EPS 1e-10
#define INF 1000000000

#define countof(a) (sizeof(a)/sizeof(a[0]))
#define PREDIACTE(t,a) [](const t & a) -> bool
#define COMPARISON_T(t) bool(*)(const t &, const t &)
#define COMPARISON(t,a,b) [](const t & a, const t & b) -> bool

//// prime ////
vector<unsigned char> isPrime;
vector<int> primes;
void initPrimes(int n)
{
	isPrime = vector<unsigned char>(n + 1, true);
	isPrime[0] = isPrime[1] = false;
	FOR(i, 2, n + 1)
	{
		if (!isPrime[i]) continue;
		primes.push_back(i);
		for (int j = i * 2; j <= n; j += i)
			isPrime[j] = false;
	}
}

//// Probability ////

// パスカルの三角形(二項定理) 2種類の並べ替えにつかう。
vector<vector<double>> makePascalTriangle(int n, bool probability = false)
{
	typedef vector<double> VD;
	vector<VD> t;
	if (!t.size()) { t.push_back(VD(1, 1)); }
	FOR(i, t.size(), n + 1)
	{
		t.push_back(VD(i + 1));
		REP(j, i)
		{
			double x = t[i - 1][j] * (probability ? 0.5 : 1);
			t[i][j] += x;
			t[i][j + 1] += x;
		}
	}
	return t;
}


//// iota iterator ////
struct iotait
{
	int n;
	iotait(int n = 0) : n(n) { }
	iotait &operator ++() { ++n; return *this; }
	int operator *() { return n; }
};

//// geo ////
struct P3
{
	double x, y, z;
	P3(double x = 0, double y = 0, double z = 0) : x(x), y(y), z(z) { }
	P3 operator +() const { return *this; }
	P3 operator +(const P3 &_) const { return P3(x + _.x, y + _.y, z + _.z); }
	P3 operator -() const { return P3(-x, -y, -z); }
	P3 operator -(const P3 &_) const { return *this + -_; }
	P3 operator *(double _) const { return P3(x*_, y*_, z*_); }
	P3 operator /(double _) const { return P3(x / _, y / _, z / _); }
	double dot(const P3 &_) const { return x*_.x + y*_.y + z*_.z; } // 内積
	P3 cross(const P3 &_) const { return P3(y*_.z - z*_.y, z*_.x - x*_.z, x*_.y - y*_.x); } // 外積
	double sqlength() const { return x*x + y*y + z*z; } // 二乗長さ
	double length() const { return sqrt(sqlength()); } // 長さ
	P3 direction() const { return *this / length(); } // 方向ベクトル
};
inline istream & operator>>(istream & in, P3 & p) { in >> p.x >> p.y >> p.z; return in; }
inline double abs(P3 p) { return p.length(); }

struct Sphere
{
	P3 c;
	double r;
	Sphere(double x, double y, double z, double r) : c(x, y, z), r(r) { }
	Sphere(P3 c, double r) : c(c), r(r) { }
	bool IntersectWith(const Sphere &rhs) const { return abs(c - rhs.c) - (r + rhs.r) < EPS; } // 接してても真。
};
inline istream & operator>>(istream & in, Sphere & c) { in >> c.c >> c.r; return in; }

struct P2
{
	double x, y;
	P2(double x = 0, double y = 0) : x(x), y(y) { }
	P2(complex<double> c) : x(c.real()), y(c.imag()) { }
	P2 operator +() const { return *this; }
	P2 operator +(const P2 &_) const { return P2(x + _.x, y + _.y); }
	P2 operator -() const { return P2(-x, -y); }
	P2 operator -(const P2 &_) const { return *this + -_; }
	P2 operator *(double _) const { return P2(x*_, y*_); }
	P2 operator /(double _) const { return P2(x / _, y / _); }
	double dot(const P2 &_) const { return x*_.x + y*_.y; } // 内積
	double cross(const P2 &_) const { return x*_.y - y*_.x; } // 外積
	double sqlength() const { return x*x + y*y; } // 二乗長さ
	double length() const { return sqrt(sqlength()); } // 長さ
	P2 orthogonal() const { return P2(y, -x); }
	P2 direction() const { return *this / length(); } // 方向ベクトル
};
inline istream & operator>>(istream & in, P2 & p) { in >> p.x >> p.y; return in; }
inline double abs(P2 p2) { return p2.length(); }
inline P2 orthogonal(P2 p) { return p.orthogonal(); }
inline complex<double> orthogonal(complex<double> c) { return  c * complex<double>(0, 1); }

// a,b から ちょうど d だけ離れた点。aとbを円周に持つ円の半径。
pair<P2, P2> get_same_distance_points(P2 a, P2 b, double d)
{
	assert(abs(a - b) <= 2 * d + EPS);
	auto v = (a + b) / 2.0 - a; // a から aとbの中点
	auto vl = abs(v);
	auto wl = sqrt(d*d - vl*vl); // 直行Vの大きさ
	auto w = orthogonal(v) * (wl / vl); // 直行V
	return make_pair(a + v + w, a + v - w);
}

struct Circle
{
	P2 c;
	double r;
	Circle(double x, double y, double r) : c(x, y), r(r) { }
	Circle(P2 c, double r) : c(c), r(r) { }
	bool IntersectWith(const Circle &rhs) const { return abs(c - rhs.c) - (r + rhs.r) < EPS; } // 接してても真。
	bool Contains(const P2 &p) const { return abs(p - c) - r < EPS; } // 接してても真。
};
inline istream & operator>>(istream & in, Circle & c) { in >> c.c >> c.r; return in; }

//// bit ////
#ifdef _MSC_VER
inline unsigned __builtin_ctz(unsigned x) { unsigned long r; _BitScanForward(&r, x); return r; }
#endif

inline int next_bit_permutation(int x)
{
	int t = x | (x - 1);
	return (t + 1) | (unsigned)((~t & -~t) - 1) >> (__builtin_ctz(x) + 1);
}

//// graph ////
struct Path
{
	int from;
	int to;
	double cost;
	Path(int from = 0, int to = 0, double cost = 0) : from(from), to(to), cost(cost) { }
	bool operator < (const Path &rhs) const { return cost < rhs.cost; }
	bool operator >(const Path &rhs) const { return cost > rhs.cost; }
};

// prim //
pair<double, vector<int>> prim(const vector<vector<double>> &costTable)
{
	int N = costTable.size();
	priority_queue<Path, vector<Path>, greater<Path>> q;
	q.push(Path(0, 0, 0));

	vector<int> parent(N, -1);
	double totalCost = 0;
	while (!q.empty())
	{
		Path cur = q.top(); q.pop();
		int i = cur.to;
		if (parent[i] != -1) continue;
		parent[i] = cur.from;
		totalCost += cur.cost;
		REP(j, N) if (parent[j] == -1) q.push(Path(i, j, costTable[i][j]));
	}
	return make_pair(totalCost, parent);
}

// dijkstra //
pair<vector<double>, vector<int>> dijkstra(const vector<vector<Path>> &routes, int start = 0, int goal = -1)
{
	int N = routes.size();
	priority_queue<Path, vector<Path>, greater<Path>> q;
	q.push(Path(start, start, 0));

	vector<int> prev(N, -1);
	vector<double> cost(N, INF);
	while (!q.empty())
	{
		Path cur = q.top(); q.pop();
		int i = cur.to;
		if (prev[i] != -1) continue;
		prev[i] = cur.from;
		cost[i] = cur.cost;
		if (i == goal) { break; }
		REP(j, routes[i].size())
		{
			Path next = Path(i, routes[i][j].to, cur.cost + routes[i][j].cost);
			if (prev[next.to] == -1)
				q.push(next);
		}
	}
	return make_pair(cost, prev);
}

//// i/o ////
template <class T>
class vevector : public vector<vector<T>>
{
public:
	vevector(int n = 0, int m = 0) : vector<vector<T>>(n, vector<T>(m)) { };
	vevector(int n, int m, const T &initial) : vector<vector<T>>(n, vector<T>(m, initial)) { };
};

template <class T> T read() { T t; cin >> t; return t; }
template <class T> vector<T> read(int n) { vector<T> v; REP(i, n) { v.push_back(read<T>()); } return v; }
template <class T> vevector<T> read(int n, int m) { vevector<T> v; REP(i, n) v.push_back(read<T>(m)); return v; }
template <class T> vevector<T> readjag(int n) { vevector<T> v; REP(i, n) v.push_back(read<T>(read<int>())); return v; }
template <class T> void write(const T &t) { cout << t << endl; }
template <class T> void write(const T &t, const T &t2) { cout << t << ' ' << t2 << endl; }
template <class T> void write(const vector<T> &v)
{
	ostringstream ss;
	for (auto x : v) ss << x << ' ';
	auto s = ss.str();
	cout << s.substr(0, s.length() - 1) << endl;
}
template<class T>
istream & operator>>(istream & in, complex<T> & n) { T r, i; in >> r >> i; n = complex<T>(r, i); return in; }

struct _Reader { template <class T> _Reader operator ,(T &rhs) { cin >> rhs; return *this; } };
#define READ(t,...) t __VA_ARGS__; _Reader(), __VA_ARGS__

template <class InIt1, class InIt2>
int partial_compare(InIt1 first1, InIt1 last1, InIt2 first2, InIt2 last2)
{
	return lexicographical_compare(first1, last1, first2, last2) ? -1
		: lexicographical_compare(first2, last2, first1, last1) ? 1
		: 0;
}

//// start up ////
void solve();
int main()
{
	// freopen("A.in", "r", stdin);
	solve();
	return 0;
}

////////////////////
/// template end ///
////////////////////

void solve()
{
	auto testcases = INF; // read<int>();
	REP(testcase, testcases)
	{
		READ(int, N);
		if (!N) { break; }

		auto ps = read<P2>(N);

		int result = min(N, 2);
		REP(i_, N) FOR(j_, i_ + 1, N)
		{
			const P2 i = ps[i_], j = ps[j_];
			double r = 1;
			if (abs(i - j) - 2 * r > -EPS) { continue; } // 遠すぎて円が作れない

			auto c = get_same_distance_points(i, j, r); // i と j から ちょうど r 離れた点

			auto c1 = Circle(c.first, r), c2 = Circle(c.second, r);
			result = max<int>(result, count_if(allof(ps), [c1, r](P2 p) { return c1.Contains(p); }));
			result = max<int>(result, count_if(allof(ps), [c2, r](P2 p) { return c2.Contains(p); }));
		}

		write(result);
	}

}