教育改革后，波利卡普在学校只学习两门科目，安全研究和体育（体育）。在第四学期的漫长几个月里，他获得了n分。当老师在日记上写一个标记时，他们没有写这个标记是针对什么科目的，他们只是写了标记。
现在是时候把日记给他严厉的父母看了。Polycarp知道，最近在家长会上，家长被告知他获得了安全研究分数和b PE分数（a + b = n）。现在，Polycarp 想在每个标记前面写一个主题的名称，以便：
正好有一个安全研究标志
正好有b PE标志
两门科目的总平均分都是最高分。
平均科目成绩是其中所有分数的总和除以它们的数量。当然，除法是以实数执行的，没有向上或向下舍入。Polycarp旨在最大化x1 + x2，其中x1是第一个科目（安全研究）的平均分数，x2是第二个科目（体育）的平均分数。
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received
n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received
a Safety Studies marks and
b PE marks (
a+b=n). Now Polycarp wants to write a subject's name in front of each mark so that:
-
there are exactly a Safety Studies marks,
-
there are exactly b PE marks,
-
the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the
x1+x2, where
x1 is the average score in the first subject (Safety Studies), and
x2 is the average score in the second one (Physical Education).
Output
Print the sequence of integers
f1,f2,...,fn, where
fi (
1≤fi≤2) is the number of a subject to which the
i-th mark should be attributed. If there are several possible solutions, then print such that the sequence
f1,f2,...,fn is the smallest lexicographically.
The sequence
p1,p2,...,pn is lexicographically less than
q1,q2,...,qn if there exists such
j (
1≤j≤n) that
pi=qi for all
1≤i<j, αnd
pj<qj.