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.DataInputStream;
import java.io.FileInputStream;
import java.io.FileWriter;
import java.io.IOException;

public class A253 {
    public static void main(String[] args) throws IOException {
        Reader scan = new Reader("input.txt");
        StringBuilder sb = new StringBuilder();

        try (FileWriter myWriter = new FileWriter("output.txt")) {
            int n = scan.nextInt();
            int m = scan.nextInt();

//            for (int i = 0; i < Math.min(n, m); i++) {
//                sb.append("BG");
//            }

            if (n < m) {
                for (int i = 0; i < n; i++) {
                sb.append("GB");
                }

                for (int i = n; i < m; i++) {
                    sb.append('G');
                }
            } else {
                for (int i = 0; i < m; i++) {
                    sb.append("BG");
                }

                for (int i = m; i < n; i++) {
                    sb.append('B');
                }
            }
            myWriter.write(String.valueOf(sb));
        } catch (Exception e) {
            System.out.print(e.toString());
        }
    }
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }
    }
}