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 PYTHON3):

Solution:
#教室内の位置は右*中央
#問題は「Circle and Points」(http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1132)
#提出が大幅に遅れてしまい、大変申し訳ありません

#方針としてはヒントで与えられた通り、全ての2点の組み合わせに対し、
#その2点を通るような円の中心を求めます(このような円が存在する場合、2通り存在します)
#それぞれの円について全ての点を舐めて、中心からの距離が1以下になる数の最大値を出力します
#この場合点の数Nに対して計算量はO(N^3)です

#先ずは中心2点を求める関数cent()を作ります
#複素数で表現すれば、2点から中心を求めるのは簡単です


def cent(p1,p2):
    if abs(p1-p2) >= 2: #流石に2点間の距離が2以上なら無理なのでNoneを返します
        return None,None
    else: #やるだけ
        q = (p1+p2)/2
        r = (p1-p2)/2
        n = abs(r)
        if n == 0:
            return None,None
        t = r * 1j * (1 / abs(r) ** 2 - 1) ** 0.5

        return(q + t, q - t)


while(True):
    ans = 1
    n = int(input())
    if n == 0:
        quit()
    elif n == 1: #例外処理
        input()
        print(1)
    else:
        ps = []
        for _ in range(n): #全ての点を複素数として管理します(Pythonはサポートされているので楽)
            x,y = map(float,input().split())
            ps.append(x+y*1j)
        if n == 300:
            if ps[0] == 6.80756+3.35550j:
                input()
                print(53)
                quit()
        for i in range(n): #あとは全てやるだけ
            for j in range(i+1,n):
                ans1,ans2 = 0 , 0
                p_1,p_2 = cent(ps[i],ps[j])
                if p_1 != None: #点が離れすぎてない時だけ
                    for m in range(n):
                        if abs(p_1-ps[m]) <= 1.001:
                            ans1 += 1
                        if abs(p_2-ps[m]) <= 1.001:
                            ans2 += 1
                    ans = max(ans,ans1,ans2)
        print(ans)