Problem: 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.

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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.

Example

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

2
5
5
11

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {

    static int n;
    static double[] x, y;
    static int ans;

    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        PrintWriter pw = new PrintWriter(System.out);

        while (true) {
            n = fs.nextInt();
            if (n == 0){
                break;
            }

            x = new double[n];
            y = new double[n];
            for(int i = 0;i < n;i++){
                x[i] = fs.nextDouble();
                y[i] = fs.nextDouble();
            }

            ans = 1;
            // 2点選ぶ全探索
            for(int i = 0;i < n;i++){
                for(int j = i+1;j<n;j++){
                    // そもそも2以上離れているなら無視
                    double dist = Math.sqrt(Math.pow(x[i] - x[j], 2) + Math.pow(y[i] - y[j], 2));
                    if(dist > 2){
                        continue;
                    }

                    double mx = (x[i] + x[j]) / 2;
                    double my = (y[i] + y[j]) / 2;
                    double m_dist = Math.sqrt(1 - (dist / 2) * (dist / 2));

                    int temp = 0;
                    double c1x = mx + (y[i] - y[j]) / dist * m_dist;
                    double c1y = my - (x[i] - x[j]) / dist * m_dist;
                    for(int k = 0;k < n;k++){
                        if(Math.pow(x[k] - c1x, 2) + Math.pow(y[k] - c1y, 2) < 1 + 1e-10){
                            temp += 1;
                        }
                    }
                    ans = Math.max(ans, temp);

                    temp = 0;
                    double c2x = mx - (y[i] - y[j]) / dist * m_dist;
                    double c2y = my + (x[i] - x[j]) / dist * m_dist;
                    for(int k = 0;k < n;k++){
                        if(Math.pow(x[k] - c2x, 2) + Math.pow(y[k] - c2y, 2) < 1 + 1e-10){
                            temp += 1;
                        }
                    }
                    ans = Math.max(ans, temp);
                }
            }

            pw.println(ans);
        }
        pw.flush();
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        public FastScanner() {
            this(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}