Innokentiy likes tea very much and today he wants to drink exactly n cups of tea. He would be happy to drink more but he had exactly n tea bags, a of them are green and b are black.

Innokentiy doesn't like to drink the same tea (green or black) more than k times in a row. Your task is to determine the order of brewing tea bags so that Innokentiy will be able to drink n cups of tea, without drinking the same tea more than k times in a row, or to inform that it is impossible. Each tea bag has to be used exactly once.

Input

The first line contains four integers n, k, a and b (1 ≤ k ≤ n ≤ 105, 0 ≤ a, b ≤ n) — the number of cups of tea Innokentiy wants to drink, the maximum number of cups of same tea he can drink in a row, the number of tea bags of green and black tea. It is guaranteed that a + b = n.

Output

If it is impossible to drink n cups of tea, print "NO" (without quotes).

Otherwise, print the string of the length n, which consists of characters 'G' and 'B'. If some character equals 'G', then the corresponding cup of tea should be green. If some character equals 'B', then the corresponding cup of tea should be black.

If there are multiple answers, print any of them.

Examples

Input

5 1 3 2


Output

GBGBG


Input

7 2 2 5


Output

BBGBGBB

Input

4 3 4 0


Output

NO
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
	public static long mod= (long) (1e9 +7);
	public static void main(String args[])
	{
		InputReader s= new InputReader(System.in);
		OutputStream outputStream= System.out;
		PrintWriter out= new PrintWriter(outputStream);
		int n= s.nextInt();
		int k= s.nextInt();
		int a= s.nextInt();
		int b= s.nextInt();
		int max = k;
	//	String ans="";
		char ans1[]= new char[n];
		//out.print(k);
		for(int i=0;i<n;i++){
			if(k==0 && i!=0){
				if(ans1[i-1]=='B'){
					ans1[i]='G';
				//	ans+='G';
					a--;
					k=max;
				}
				else{
					ans1[i]='B';
				//	ans+='B';
					b--;
					k=max;
				}
				continue;
			}
			int x = Math.max(a, b);
			if(x==a){
				ans1[i]= 'G';
	//				ans+='G';
					k--;
					a--;
			}
			else{
				ans1[i]= 'B';
		//		ans+='B';
				k--;
				b--;
			}
		}
		if(a<0 || b<0)
			out.println("NO");
		else{
			for(int i=0;i<n;i++){
				out.print(ans1[i]);
			}
		}
		//	out.println(ans);
		out.close();
	}
	
	static long combinations(long n,long r){   // O(r)
		if(r==0 || r==n) return 1;
		r= Math.min(r, n-r);
		long ans=n;  // nC1=n;
		for(int i=1;i<r;i++){
			ans= ans*(n-i);
			ans= ans/(i+1);
		}
		return ans;
		
	}
	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}
	static long modulo(long a,long b,long c) {
		long x=1;
		long y=a;
		while(b > 0){
			if(b%2 == 1){
				x=(x*y)%c;
			}
			y = (y*y)%c; // squaring the base
			b /= 2;
		}
		return  x%c;
	}
	static void catalan_numbers(int n) {
		long catalan[]= new long[n+1];
	    catalan[1] = 1;
	    for (int i = 2; i <= n; i++) {
	        for (int j = 1; j <= i - 1; j++) {
	            catalan[i] = catalan[i] + ((catalan[j]) * catalan[i - j]);
	        }
	    }
	}
	
	static ArrayList<Integer> primeFactors(int n)   // O(sqrt(n))
	{
	    // Print the number of 2s that divide n
		ArrayList<Integer> arr= new ArrayList<>();
	    while (n%2 == 0)
	    {
	        arr.add(2);
	        n = n/2;
	    }
	 
	    // n must be odd at this point.  So we can skip one element (Note i = i +2)
	    for (int i = 3; i <= Math.sqrt(n); i = i+2)
	    {
	        // While i divides n, print i and divide n
	        while (n%i == 0)
	        {
	            arr.add(i);
	            n = n/i;
	        }
	    }
	    // This condition is to handle the case when n is a prime number
	    // greater than 2
	    if (n > 2)
	        arr.add(n);
	    
	    return arr;
	}

	public static long expo(long a, long b){
		if (b==0)
	        return 1%mod;
	    if (b==1)
	        return a%mod;
	    if (b==2)
	        return ((a%mod)*(a%mod))%mod;

	    if (b%2==0){
	            return expo(expo(a%mod,(b%mod)/2)%mod,2%mod)%mod;
	    }
	    else{
	        return (a%mod)*(expo(expo(a%mod,(b-1)%mod/2)%mod,2%mod)%mod)%mod;
	    }
	}
	static class Pair implements Comparable<Pair>
	{
		long f;
		String s;
		Pair(long ii, String cc)
		{
			f=ii;
			s=cc;
		}
		
		public int compareTo(Pair o) 
		{
			return Long.compare(this.f, o.f);
		}
		
	}
	
	public static int[] sieve(int N){    // O(n*log(logn))
		int arr[]= new int[N+1];
		for(int i=2;i<=Math.sqrt(N);i++){
			if(arr[i]==0){
				for(int j= i*i;j<= N;j= j+i){
					arr[j]=1;
				}
			}
		}
		return arr;
		// All the i for which arr[i]==0 are prime numbers.
	}
	static long gcd(long a,long b){       // O(logn)
		if(b==0) return a;
		return gcd(b,a%b);
	}
	static class InputReader {

	    public BufferedReader reader;
	    public StringTokenizer tokenizer;
 
	    public InputReader(InputStream inputstream) {
	      reader = new BufferedReader(new InputStreamReader(inputstream));
	      tokenizer = null;
	    }
	    
	    public String nextLine(){
	    	String fullLine=null;
	    	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	              fullLine=reader.readLine();
	            } catch (IOException e) {
	              throw new RuntimeException(e);
	            }
	            return fullLine;
	          }
	          return fullLine;
	    }
		public String next() {
	      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	        try {
	          tokenizer = new StringTokenizer(reader.readLine());
	        } catch (IOException e) {
	          throw new RuntimeException(e);
	        }
	      }
	      return tokenizer.nextToken();
	    }
		public long nextLong() {
		      return Long.parseLong(next());
		    }
	    public int nextInt() {
	      return Integer.parseInt(next());
	    }
	  }
}