CF351A.Jeff and Rounding

Jeff got 2_n_ real numbers _a_1, _a_2, ..., a_2_n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows:

• choose indexes i and j (i ≠ j) that haven't been chosen yet;
• round element a__i to the nearest integer that isn't more than a__i (assign to a__i: ⌊ a__i ⌋);
• round element a__j to the nearest integer that isn't less than a__j (assign to a__j: ⌈ a__j ⌉).

Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference.

杰夫收到了 $2n$ 个实数 $a_1, a_2, \dots, a_{2n}$ 作为生日礼物。这个男孩讨厌非整数，因此他决定对收到的数字稍作“调整”。具体来说，杰夫依次执行 $n$ 次操作，每次操作如下：

• 选择两个尚未被选过的下标 $i$ 和 $j$（其中 $i \neq j$）；
• 将元素 $a_i$ 向下取整（即赋值为 $a_i := \lfloor a_i \rfloor$）；
• 将元素 $a_j$ 向上取整（即赋值为 $a_j := \lceil a_j \rceil$）。

然而，杰夫不想伤害送他这个序列的人的感情。因此，他希望以某种方式执行这些操作，使得操作前所有元素之和与操作后所有元素之和的绝对差值尽可能小。请帮助杰夫找出该绝对差值的最小可能值。

The first line contains integer n (1 ≤ n ≤ 2000). The next line contains 2_n_ real numbers _a_1, _a_2, ..., a_2_n (0 ≤ a__i ≤ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces.

第一行包含一个整数 $n$（$1 \leq n \leq 2000$）。第二行包含 $2n$ 个实数 $a_1, a_2, \dots, a_{2n}$（$0 \leq a_i \leq 10000$），每个数均精确到小数点后三位。这些数以空格分隔。

In a single line print a single real number — the required difference with exactly three digits after the decimal point.

在一行中输出一个实数——所要求的差值，小数点后精确保留三位数字。

In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.

在第一个测试用例中，你需要执行如下操作：$(i = 1,\, j = 4)$、$(i = 2,\, j = 3)$、$(i = 5,\, j = 6)$。此时，差值为 $|(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25$。