CodeForces 253A Boys and Girls

Description

input

input.txt

output

output.txt

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.

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

Input

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

Output

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.

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.

Sample Input

Input

3 3

Output

GBGBGB

Input

4 2

Output

BGBGBB

Hint

In the first sample another possible answer is BGBGBG.

In the second sample answer BBGBGB is also optimal.

题意：有N个男的M个女的站成一排，输出男女相邻的数量最大的排列方式。

水题。。。

#include<stdio.h>
#include<queue>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<assert.h>
#include<stdlib.h>
#include<time.h>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<fstream>
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
#define debug printf("---------------\n");
using namespace std;

int main()
{
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    int n,m;
    scanf("%d%d",&n,&m);
    if(n>m)
    {
        while(n||m)
        {
            if(n)
            {
                printf("B");
                n--;
            }
            if(m)
            {
                printf("G");
                m--;
            }
        }
    }
    else
    {
        while(n||m)
        {
            if(m)
            {
                printf("G");
                m--;
            }
            if(n)
            {
                printf("B");
                n--;
            }

        }
    }
    return 0;
}