O - Ciel and Robot 模拟

O - Ciel and Robot

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:  System Crawler  (2013-07-04)

Description

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 strings. 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.

Sample Input

Input

2 2
RU

Output

Yes

Input

1 2
RU

Output

No

Input

-1 1000000000
LRRLU

Output

Yes

Input

0 0
D

Output

Yes

const int maxn = 300;
char op[maxn];
struct Point
{
    int x;
    int y;
};
vector<Point> p;
int main() 
{
    //freopen("in.txt","r",stdin);
    int x,y;
    while(cin>>x>>y)
    {
        p.clear();
        scanf("%s",op);
        int len = strlen(op);
        int X = 0;
        int Y = 0;
        p.push_back((Point){0,0});
        rep(i,0,len)
        {
            if(op[i] == 'L') X--;
            if(op[i] == 'R') X++;
            if(op[i] == 'U') Y++;
            if(op[i] == 'D') Y--;
            p.push_back((Point){X,Y});
        }
        int flag = 0;
        rep(i,0,p.size())
        {
            if(X == 0 && Y)
            {
                if(x == p[i].x &&  (y - p[i].y)%Y == 0 && (y - p[i].y)/Y >= 0)
                {
                    flag = 1;
                    break;
                }
                continue;
            }
            if(Y == 0 && X)
            {
                if(y == p[i].y && (x - p[i].x)%X == 0 && (x - p[i].x)/X >= 0)
                {
                     flag = 1;
                     break;
                }
                continue;
            }
            if(X == 0 && Y == 0)
            {
               if( x == p[i].x && y == p[i].y)
               {
                   flag = 1;
                   break;
               }
               continue;
            }
            if((x - p[i].x)%X == 0 &&  (y - p[i].y)%Y == 0 && ((x - p[i].x)/X == (y - p[i].y)/Y) && (x - p[i].x)/X >= 0)
            {
                flag = 1;
                break;
            }
        }
        if(flag)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}