p00708 Building a Space Station
You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a computer program to complete the task.

The space station is made up with a number of units, called cells. All cells are sphere-shaped, but their sizes are not necessarily uniform. Each cell is fixed at its predetermined position shortly after the station is successfully put into its orbit. It is quite strange that two cells may be touching each other, or even may be overlapping. In an extreme case, a cell may be totally enclosing another one. I do not know how such arrangements are possible.

All the cells must be connected, since crew members should be able to walk from any cell to any other cell. They can walk from a cell A to another cell B, if, (1) A and B are touching each other or overlapping, (2) A and B are connected by a `corridor', or (3) there is a cell C such that walking from A to C, and also from B to C are both possible. Note that the condition (3) should be interpreted transitively.

You are expected to design a configuration, namely, which pairs of cells are to be connected with corridors. There is some freedom in the corridor configuration. For example, if there are three cells A, B and C, not touching nor overlapping each other, at least three plans are possible in order to connect all three cells. The first is to build corridors A-B and A-C, the second B-C and B-A, the third C-A and C-B. The cost of building a corridor is proportional to its length. Therefore, you should choose a plan with the shortest total length of the corridors.

You can ignore the width of a corridor. A corridor is built between points on two cells' surfaces. It can be made arbitrarily long, but of course the shortest one is chosen. Even if two corridors A-B and C-D intersect in space, they are not considered to form a connection path between (for example) A and C. In other words, you may consider that two corridors never intersect.



Input

The input consists of multiple data sets. Each data set is given in the following format.

> n
>  x1 y1 z1 r1
>  x2 y2 z2 r2
>  ...
>  xn yn zn rn
>

The first line of a data set contains an integer n, which is the number of cells. n is positive, and does not exceed 100.

The following n lines are descriptions of cells. Four values in a line are x-, y- and z-coordinates of the center, and radius (called r in the rest of the problem) of the sphere, in this order. Each value is given by a decimal fraction, with 3 digits after the decimal point. Values are separated by a space character.

Each of x, y, z and r is positive and is less than 100.0.

The end of the input is indicated by a line containing a zero.

Output

For each data set, the shortest total length of the corridors should be printed, each in a separate line. The printed values should have 3 digits after the decimal point. They may not have an error greater than 0.001.

Note that if no corridors are necessary, that is, if all the cells are connected without corridors, the shortest total length of the corridors is 0.000.

Example

Input

3
10.000 10.000 50.000 10.000
40.000 10.000 50.000 10.000
40.000 40.000 50.000 10.000
2
30.000 30.000 30.000 20.000
40.000 40.000 40.000 20.000
5
5.729 15.143 3.996 25.837
6.013 14.372 4.818 10.671
80.115 63.292 84.477 15.120
64.095 80.924 70.029 14.881
39.472 85.116 71.369 5.553
0


Output

20.000
0.000
73.834
```python
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1127
# クラスカル or プリム法

import math


class UnionFind:
    def __init__(self, n):
        self.n = n
        # 各要素の親要素の番号を格納するリスト
        # 根がルートの時は、グループの要素数を格納　最初は全て根で要素数1
        self.parents = [-1] * n

    # 再帰 groupの根を返す
    # 経路圧縮あり
    def find(self, x):
        # 0より小さい場合は根
        if self.parents[x] < 0:
            # indexを返す
            return x
        else:
            # 根になるまで親を辿る
            # 経路圧縮 根のindexが全ての親に再帰的につく
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    # groupを併合
    def union(self, x, y):
        # 根を探す
        x = self.find(x)
        y = self.find(y)

        # 最初から統合されている
        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x
        # 根のグループ数を追加 xの方グループの要素数が多い前提なので多い方に寄せる
        self.parents[x] += self.parents[y]
        # 親(根)のindexを入れる
        self.parents[y] = x

    # xのgroupのサイズ
    def size(self, x):
        return - self.parents[self.find(x)]

    # 根が同じか判定
    def same(self, x, y):
        return self.find(x) == self.find(y)

    # ルートが同じ要素を返す
    def members(self, x):
        root = self.find(x)
        # 根が同じ要素
        return [i for i, in range(self.n) if self.find(i) == root]

    # 全ての根の要素を返す
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    # グループの数を数える 遅い
    def group_count(self):
        return len(self.roots())

    # {root: [要素]}
    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())


while True:
    n = int(input())
    uf = UnionFind(n)
    if n == 0:
        break
    cell = []
    for _ in range(n):
        x, y, z, r = map(float, input().split())
        cell.append((x, y, z, r))

    edge = []
    for i in range(n):
        for j in range(i + 1, n):
            dis = math.sqrt(
                (cell[i][0] - cell[j][0]) ** 2 + (cell[i][1] - cell[j][1]) ** 2 + (cell[i][2] - cell[j][2]) ** 2)
            #print(dis)
            if dis < (cell[i][3] + cell[j][3]):
                uf.union(i, j)
            else:
                edge.append((dis - (cell[i][3] + cell[j][3]), i, j))
    edge.sort()
    ans = 0
    num = n
    for c, a, b in edge:
        # 同じグループでなかったら統合
        if not uf.same(a, b):
            uf.union(a, b)
            ans += c
            num -= 1
            # グループ数は開催する都市の数になったら終了
            if num == 1:
                break
    print('{:.3f}'.format(ans))

```