Problem: 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.

Examples

Input

3 3


Output

GBGBGB


Input

4 2


Output

BGBGBB

Note

In the first sample another possible answer is BGBGBG. 

In the second sample answer BBGBGB is also optimal.

Solution (in JAVA):

Solution:
/*input
3 3

*/
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	static PrintWriter out;
	static int MOD = 1000000007;
	static FastReader scan;

    /*-------- I/O using short named function ---------*/
	public static String ns(){return scan.next();}
	public static int ni(){return scan.nextInt();}
	public static long nl(){return scan.nextLong();}
	public static double nd(){return scan.nextDouble();}
	public static String nln(){return scan.nextLine();}
	public static void p(Object o){out.print(o);}
	public static void ps(Object o){out.print(o + " ");}
	public static void pn(Object o){out.println(o);}
    /*-------- for output of an array ---------------------*/
     static void iPA(int arr []){
     	 StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void lPA(long arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void sPA(String arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void dPA(double arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
    /*-------------- for input in an array ---------------------*/
	static void iIA(int arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ni();
	}
	static void lIA(long arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nl(); 
	}
	static void sIA(String arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ns(); 
	}
	static void dIA(double arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nd();
	}
	/*------------ for taking input faster ----------------*/
     static class FastReader
	{
		BufferedReader br;
		StringTokenizer st;

		public FastReader()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next()
		{
			while (st == null || !st.hasMoreElements())
			{
				try
				{
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e)
				{
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt()
		{
			return Integer.parseInt(next());
		}

		long nextLong()
		{
			return Long.parseLong(next());
		}

		double nextDouble()
		{
			return Double.parseDouble(next());
		}

		String nextLine()
		{
			String str = "";
			try
			{
				str = br.readLine();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}
			return str;
		}
	}

	// Method to check if x is power of 2
    static boolean isPowerOfTwo (int x) { return x!=0 && ((x&(x-1)) == 0);}
    //Method to return lcm of two numbers
    static int gcd(int a, int b){return a==0?b:gcd(b % a, a); }
    //Method to count digit of a number
    static int countDigit(long n){return (int)Math.floor(Math.log10(n) + 1);}

    //Method for sorting
	static void ruffle_sort(int[] a) {
		//shandom_ruffle
		Random r=new Random();
		int n=a.length;
		for (int i=0; i<n; i++) {
			int oi=r.nextInt(n);
			int temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		//sort
		Arrays.sort(a);
	}
	//Method for checking if a number is prime or not
	static boolean isPrime(int n) { 
        if (n <= 1) return false; 
        if (n <= 3) return true; 
        if (n % 2 == 0 || n % 3 == 0) return false; 
        for (int i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
        return true; 
    } 

	public static void main (String[] args) throws java.lang.Exception
	{
		OutputStream outputStream =System.out;
        out =new PrintWriter("output.txt");
        scan =new FastReader();
		//for fast output sometimes
		StringBuilder sb = new StringBuilder();
		Scanner scan = new Scanner(new File("input.txt"));
		int n = scan.nextInt(), m = scan.nextInt();
		char list[] = new char[m+n];
		Arrays.fill(list, 'a');
		int i=1;
		if(n<m){
			for(int k=0; k<n; k++){
				list[i]='B';
				i+=2;
			}	
			for(int j=0; j<m+n; j++){
				if(list[j]!='B')
					list[j]='G';
			}
		}else{
			for(int k=0; k<m; k++){
				list[i]='G';
				i+=2;
			}
			for(int j=0; j<m+n; j++){
				if(list[j]!='G')
					list[j]='B';
			}
		}
		for(int j=0; j<m+n; j++){
			p(list[j]);
		}
		pn("");
        out.flush();
		out.close();
	}
	
}