We have a rectangular grid of squares with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left. On this grid, there is a piece, which is initially placed at square (s_r,s_c).

Takahashi and Aoki will play a game, where each player has a string of length N. Takahashi's string is S, and Aoki's string is T. S and T both consist of four kinds of letters: `L`, `R`, `U` and `D`.

The game consists of N steps. The i-th step proceeds as follows:

* First, Takahashi performs a move. He either moves the piece in the direction of S_i, or does not move the piece.
* Second, Aoki performs a move. He either moves the piece in the direction of T_i, or does not move the piece.



Here, to move the piece in the direction of `L`, `R`, `U` and `D`, is to move the piece from square (r,c) to square (r,c-1), (r,c+1), (r-1,c) and (r+1,c), respectively. If the destination square does not exist, the piece is removed from the grid, and the game ends, even if less than N steps are done.

Takahashi wants to remove the piece from the grid in one of the N steps. Aoki, on the other hand, wants to finish the N steps with the piece remaining on the grid. Determine if the piece will remain on the grid at the end of the game when both players play optimally.

Constraints

* 2 \leq H,W \leq 2 \times 10^5
* 2 \leq N \leq 2 \times 10^5
* 1 \leq s_r \leq H
* 1 \leq s_c \leq W
* |S|=|T|=N
* S and T consists of the four kinds of letters `L`, `R`, `U` and `D`.

Input

Input is given from Standard Input in the following format:


H W N
s_r s_c
S
T


Output

If the piece will remain on the grid at the end of the game, print `YES`; otherwise, print `NO`.

Examples

Input

2 3 3
2 2
RRL
LUD


Output

YES


Input

4 3 5
2 2
UDRRR
LLDUD


Output

NO


Input

5 6 11
2 1
RLDRRUDDLRL
URRDRLLDLRD


Output

NO
#include <cstdio>
using namespace std;

int h, w, n;
int sr, sc;
char t[200001];
char a[200001];
int d[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };
char dir[5] = "RLDU";

int main()
{
	scanf("%d %d %d", &h, &w, &n);
	scanf("%d %d", &sr, &sc);
	scanf("%s", t);
	scanf("%s", a);

	int by, bx;
	int flag = 0;

	for (int i = 0; i < 4; i++)
	{
		by = sr;
		bx = sc;
		int temp = i - (i % 2) * 2 + 1;

		for (int j = 0; j < n && flag == 0; j++)
		{
			if (t[j] == dir[i])
			{
				by += d[i][0];
				bx += d[i][1];
			}
			if (by > h || by < 1 || bx > w || bx < 1)
			{
				flag = 1;
				break;
			}		

			if (a[j] == dir[temp])
			{
				if (by + d[temp][0] > h || by + d[temp][0] < 1 || bx + d[temp][1] > w || bx + d[temp][1] < 1)
					continue;

				by += d[temp][0];
				bx += d[temp][1];
			}
		}
	}
	
	if (flag == 0)
		printf("YES");
	else
		printf("NO");

	return 0;
}