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:
import java.io.*;
import java.math.BigInteger;
import java.util.*;



public class Boys_and_Girls {

	static int x[],memo[][],ans,a[];
	static int n,p[],w[],W;
	static HashSet<Integer> hs[];
	static StringBuilder sb=new StringBuilder();
	public static void main(String[] args) throws Exception{
		Scanner sc=new Scanner("input.txt");
		PrintWriter out=new PrintWriter("output.txt");

		//Scanner sc=new Scanner(System.in);
		int B=sc.nextInt();
		int G=sc.nextInt();
		if(B>G)Bf(B,G);
		else Gf(B,G);
		out.println(sb);


		out.flush();
	}
	static void Bf(int B,int G) {
		while(B>0||G>0) {
			if (B>0) {
				sb.append("B");
				B--;
			}

			if (G>0) {
				sb.append("G");
				G--;
			}


		}
	}
	static void Gf(int B,int G) {
		while(B>0||G>0) {
			if (G>0) {
				sb.append("G");
				G--;
			}

			if (B>0) {
				sb.append("B");
				B--;
			}


		}
	}
	static boolean [] vis,vis2;
	static void dfs(int u) {
		vis[u]=true;
		for(int v:hs[u]) {
			if(!vis[v]) {
				dfs(v);
			}
		}
	}
	static void dfs2(int u) {
		vis2[u]=true;
		for(int v:hs[u]) {
			if(!vis[v]) {
				dfs(v);
			}
		}
	}
	static class hi implements Comparable<hi>{
		int node;
		int val;

		hi(int n,int v){
			node=n;
			val=v;
		}

		@Override
		public int compareTo(hi o) {
			return o.val-val;
		}
	}
	static int dp(int i,int sumsofar) {
		//System.out.println("##");
		if(i==n) {
			//	System.out.println(sumsofar);
			if(sumsofar%360==0) return 1;
			else return 0;
		}
		if(memo[i][sumsofar]!=-1) {
			//	out.println("HI");
			return memo[i][sumsofar];
		}//System.out.println((sumsofar+a[i]+360)%360+" "+(sumsofar-a[i]+360)%360+" "+i);
		int take=dp(i+1,(sumsofar+a[i]+360)%360);
		int leave=dp(i+1,(sumsofar-a[i]+360)%360);

		return memo[i][sumsofar]=take|leave;


	}

	static boolean valid(int x,int n) {
		return x>=0&&x<n;
	}






	static class mando implements Comparable<mando>{
		char c;
		int val;
		mando(char cc,int v){
			c=cc;
			val=v;
		}
		@Override
		public int compareTo(mando o) {
			return val-o.val;
		}

	}
	static int []visited;
	static class Scanner{
		StringTokenizer st;
		BufferedReader br;
		public Scanner(InputStream system) { br = new BufferedReader(new InputStreamReader(system));}
		public Scanner(String file) throws Exception{br = new BufferedReader(new FileReader (file));}
		public String next() throws IOException{
			while (st==null|| !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int nextInt() throws IOException{
			return Integer.parseInt(next());
		}
		public long nextLong() throws IOException{
			return Long.parseLong(next());
		}
		public String nextLine() throws IOException{
			return br.readLine();
		}


	}

}