Fox Ciel has a robot on a 2D plane. Initially it is located in (0, 0). Fox Ciel code a command to it. The command was represented by string s. Each character of s is one move operation. There are four move operations at all:

  * 'U': go up, (x, y)  →  (x, y+1); 
  * 'D': go down, (x, y)  →  (x, y-1); 
  * 'L': go left, (x, y)  →  (x-1, y); 
  * 'R': go right, (x, y)  →  (x+1, y). 



The robot will do the operations in s from left to right, and repeat it infinite times. Help Fox Ciel to determine if after some steps the robot will located in (a, b).

Input

The first line contains two integers a and b, ( - 109 ≤ a, b ≤ 109). The second line contains a string s (1 ≤ |s| ≤ 100, s only contains characters 'U', 'D', 'L', 'R') — the command.

Output

Print "Yes" if the robot will be located at (a, b), and "No" otherwise.

Examples

Input

2 2
RU


Output

Yes


Input

1 2
RU


Output

No


Input

-1 1000000000
LRRLU


Output

Yes


Input

0 0
D


Output

Yes

Note

In the first and second test case, command string is "RU", so the robot will go right, then go up, then right, and then up and so on.

The locations of its moves are (0, 0)  →  (1, 0)  →  (1, 1)  →  (2, 1)  →  (2, 2)  →  ...

So it can reach (2, 2) but not (1, 2).
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;
import java.util.TreeSet;


public class Main {

	/**
	 * @param args
	 */
	 BufferedReader in;
	    PrintWriter out = new PrintWriter(System.out);
	    StringTokenizer tok;
	   
	     String readToken() throws IOException { // reads the token; to read a full line use in.readLine()
	    while (tok == null || !tok.hasMoreTokens()) {
	    tok = new StringTokenizer(in.readLine());
	    	
	    }
	    return tok.nextToken(); // sometimes it's better to use nextToken(String) method here
	    }
	     int readInt() throws IOException { // write readDouble(), readLong(), readBigInteger() methods if necessary
	    return Integer.parseInt(readToken());
	    }
	     public String readLine() throws IOException{
			return in.readLine();
	    	 
	     }
	     long readLong() throws IOException { // write readDouble(), readLong(), readBigInteger() methods if necessary
	    	    return Long.parseLong(readToken());
	    	    }
	     double readDouble() throws IOException { // write readDouble(), readLong(), readBigInteger() methods if necessary
	    	    return Double.parseDouble(readToken());
	    	    }
	    public static void main(String[] args) throws IOException
	    {
	       new Main().run();
	    }
	    void run() throws IOException
	    {
	    	boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	    	out = new PrintWriter(new OutputStreamWriter(System.out));
	    	in = new BufferedReader(new InputStreamReader(System.in));
	    	if (oj){
	    
	    	}
	    	else{
	    		
	    		
	    	}
	       solve();
	       out.flush();
	       tok=null;
	    }
	  
	    
	    void solve() throws IOException{
	    	/*mvX[i]: thesi X meta apo ta prwta i bimata tis akolouthias
	    	 * k.o.k mvY
	    	 * arxika, meta to telos tis seiras
	    	 * o robot hta einai stin thesi (x+mvX[n-1],y+mvY[n-1])
	    	 * kai antistoxia meta apo k epanalipseis
	    	 *(x+kmvX[n-1],y+kmvY[n-1])
	    	 *opote auto pou kanw einai na kanw mia diapreasi tou array
	    	 *kai  koitaw an isxei to eksis:
	    	 *kmvX[n-1]+mvX[i]==x && kmvY[n-1]+mvY[i]==y
	    	 *gia na brw to k linw mia ek twn dio eksiswseis
	    	 *epeidi to mvX[n-1]*k einai gramiki eksiswsi ehei mia lisi gia kathe timi x.
	    	 *Opote, an to brw ehei kalws! 
	    	 */
	    	int x=readInt();
	    	int y=readInt();
	    	if (x==0 && y==0){
	    		out.println("Yes");
	    	}
	    	else{
	    	String s=readLine();
	    	int n=s.length();
	    	int[] mvX=new int[n];
	    	int[] mvY=new int[n];
	    	for (int i=0;i<n;i++){
	    		char c=s.charAt(i);
	    		if (c=='U'){
	    			if (i==0){
	    				mvX[0]=0;
	    				mvY[0]=1;
	    			}else{
	    				mvX[i]=mvX[i-1];
	    				mvY[i]=mvY[i-1]+1;
	    			}
	    		}else if (c=='D'){
	    			if (i==0){
	    				mvX[0]=0;
	    				mvY[0]=-1;
	    			}else{
	    				mvX[i]=mvX[i-1];
	    				mvY[i]=mvY[i-1]-1;
	    			}
	    		}else if (c=='R'){
	    			if (i==0){
	    				mvY[0]=0;
	    				mvX[0]=1;
	    			}else{
	    				mvY[i]=mvY[i-1];
	    				mvX[i]=mvX[i-1]+1;
	    			}
	    		}else{
	    			if (i==0){
	    				mvY[0]=0;
	    				mvX[0]=-1;
	    			}else{
	    				mvY[i]=mvY[i-1];
	    				mvX[i]=mvX[i-1]-1;
	    			}
	    		}
	    	}
	    	boolean hasSol=false;
	    	for (int i=0;i<n;i++){
	    		if (mvX[n-1]!=0){
	    			int k=(x-mvX[i])/mvX[n-1];
	    			if (k>=0){
	    				if (k*mvX[n-1]+mvX[i]==x){ //integer check!
	    					if (k*mvY[n-1]+mvY[i]==y){
	    						hasSol=true;
	    						break;
	    					}
	    				}
	    			}
	    		}else{
	    			if (mvY[n-1]!=0){
	    				int k=(y-mvY[i])/mvY[n-1];
	    				
		    			if (k>=0){
		    				if (k*mvY[n-1]+mvY[i]==y){ //integer check!
		    					if (k*mvX[n-1]+mvX[i]==x){
		    						hasSol=true;
		    						break;
		    					}
		    				}
		    			}
	    			}else{
	    				if (x==mvX[i] && y==mvY[i]){
	    					hasSol=true;
	    					break;
	    				}
	    			}
	    		}
	    	}
	    	if (hasSol){
	    		out.println("Yes");
	    	}else{
	    		out.println("No");
	    	}
	    }
	    }
	    
}