題目：

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

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).

解題報告：比較水的一道題目，難點就是不斷的迴圈尋找每次遍歷會達到的點，進行判斷。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;

const int maxn =1500;

int main()
{
	ll A,B;
	char str[maxn];
	while(scanf("%I64d%I64d",&A,&B)!=EOF)
	{
		scanf("%s",str);
		ll n=strlen(str);
		ll xx,yy;
		xx=0;yy=0;
		int flag=0;
		int tx=0,ty=0;
		for(int i=0;i<100000;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(tx==A&&ty==B) flag=1;
				if(str[j]=='U') ty++;
				if(str[j]=='D') ty--;
				if(str[j]=='L') tx--;
				if(str[j]=='R') tx++;
				if(tx==A&&ty==B) flag=1;	
			}
		}
			for(ll i=0;i<n;i++)
			{
				if(xx==A&&yy==B) flag=1;
				if(str[i]=='U') yy++;
				if(str[i]=='D') yy--;
				if(str[i]=='L') xx--;
				if(str[i]=='R') xx++;
				if(xx==A&&yy==B) flag=1;
			}
			if(flag==1)
			{
				printf("Yes\n");
				continue;
			}
			if(xx==yy&&xx==0)
			{
				printf("No\n");
				continue;
			}
			if(A*xx<0||B*yy<0)
			{
				printf("No\n");
				continue;
			}
			ll x=0,y=0;
			for(ll i=0;i<n;i++)
			{
				if((A-x)*yy==(B-y)*xx)
				{
					if(xx!=0&&(A-x)%xx==0)
					{
						flag=1;
					}
					if(yy!=0&&(B-y)%yy==0)
					{
						flag=1;
					}
				}
				if(str[i]=='U') y++;
				if(str[i]=='D') y--;
				if(str[i]=='L') x--;
				if(str[i]=='R') x++; 
				if((A-x)*yy==(B-y)*xx)
				{
					if(xx!=0&&(A-x)%xx==0)
					{
						flag=1;
					}
					if(yy!=0&&(B-y)%yy==0)
					{
						flag=1;
					}
				}
			}	
		 
		if(flag==1) printf("Yes\n");
		else printf("No\n");
	}	
}