Codeforces Round 154 (Div. 2)
A
Boys and Girls
Codeforces
1100
原题
←B →
翻译中…

There are $*n*$ boys and $*m*$ girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to $*n* + *m*$. Then the number of integers $*i*$ ($1 ≤ *i* < *n* + *m*$) such that positions with indexes $*i*$ and $*i* + 1$ contain children of different genders (position $*i*$ has a girl and position $*i* + 1$ has a boy or vice versa) must be as large as possible.

翻译中…

Help the children and tell them how to form the line.

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Input

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The single line of the input contains two integers $*n*$ and $*m*$ ($1 ≤ *n*, *m* ≤ 100$), separated by a space.

翻译中…

Output

翻译中…

Print a line of $*n* + *m*$ characters. Print on the $*i*$-th position of the line character "B", if the $*i*$-th position of your arrangement should have a boy and "G", if it should have a girl.

翻译中…

Of course, the number of characters "B" should equal $*n*$ and the number of characters "G" should equal $*m*$. If there are multiple optimal solutions, print any of them.

翻译中…

Note

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In the first sample another possible answer is BGBGBG.

翻译中…

In the second sample answer BBGBGB is also optimal.

翻译中…

样例

Input #1
3 3
Output #1
GBGBGB
InputCopy OutputCopy
Input #2
4 2
Output #2
BGBGBB
InputCopy OutputCopy
题目信息
题号
253A
时间限制
1 s
内存限制
256 MB
标签
贪心

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。