Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 3 3
2 2
RRL
LUD

Sample Output 1

YES

Here is one possible progress of the game:

Takahashi moves the piece right. The piece is now at (2,3).

Aoki moves the piece left. The piece is now at (2,2).

Takahashi does not move the piece. The piece remains at (2,2).

Aoki moves the piece up. The piece is now at (1,2).

Takahashi moves the piece left. The piece is now at (1,1).

Aoki does not move the piece. The piece remains at (1,1).

Sample Input 2

4 3 5
2 2
UDRRR
LLDUD

Sample Output 2

NO

Sample Input 3

5 6 11
2 1
RLDRRUDDLRL
URRDRLLDLRD

Sample Output 3

NO

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.HashMap;
import java.util.Map;

/**
 * Copyright © 2018 Chris. All rights reserved.
 *
 * @author Chris
 * 2018/7/9 15:33
 * @see format
 */
public class Main {

    private static BufferedReader br;
    private static StreamTokenizer st;
    private static PrintWriter pw;

    static final int INF = 1000000007;
    static int MOD = 1000000007;

    static int[] getBit(long num) {
        int bit[] = new int[40];
        int cur = 0;
        while (num > 0) {
            bit[cur++] = (int) (num & 1);
            num >>= 1;
        }
        return bit;
    }

    static Map<Integer, Integer> dddd = new HashMap<>();

    private static void solve() throws IOException {
        int H = nextInt();
        int W = nextInt();
        int N = nextInt();

        int r = nextInt();
        int c = nextInt();
        char S[] = next(N).toCharArray();
        char T[] = next(N).toCharArray();

        Map<Character, Integer> map = new HashMap<>();
        map.put('L', 0);
        map.put('R', 1);
        map.put('U', 2);
        map.put('D', 3);
        boolean ok = true;
        int max[][] = new int[2][4];
        int cnt[] = {c - 1, W - c, r - 1, H - r};
        for (int i = 0; i < N; i++) {
            int idx = map.get(S[i]);
            int tar = idx + 1;
            if (tar % 2 == 0) {
                tar -= 2;
            }
            max[0][idx]++;
            if (max[0][idx] - max[1][tar] > cnt[idx]
                    || max[0][idx] - cnt[idx] > cnt[tar]) {
                ok = false;
                break;
            }

            idx = map.get(T[i]);
            tar = idx + 1;
            if (tar % 2 == 0) {
                tar -= 2;
            }
            max[1][idx]++;
            int min = Math.min(max[0][tar], max[1][idx]);
            max[0][tar] -= min;
            max[1][idx] -= min;
        }
        if (ok) {
            pw.println("YES");
        } else {
            pw.println("NO");
        }
    }

    static void swap(long a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void swap(int a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void getDiv(Map<Integer, Integer> map, int n) {
        int sqrt = (int) Math.sqrt(n);
        for (int i = sqrt; i >= 2; i--) {
            if (n % i == 0) {
                getDiv(map, i);
                getDiv(map, n / i);
                return;
            }
        }
        map.put(n, map.getOrDefault(n, 0) + 1);
    }

    public static boolean[] generatePrime(int n) {
        boolean p[] = new boolean[n + 1];
        p[2] = true;

        for (int i = 3; i <= n; i += 2) {
            p[i] = true;
        }

        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (!p[i]) {
                continue;
            }
            for (int j = i * i; j <= n; j += i << 1) {
                p[j] = false;
            }
        }
        return p;
    }

    static long llMod(long a, long b, long mod) {
        return (a * b - (long) ((double) a / mod * b + 0.5) * mod + mod) % mod;
    }

    static int pow(long a, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = (ans * a) % MOD;
            }
            a = (a * a) % MOD;
            n >>= 1;
        }
        return (int) ans;
    }

    static int pow(long a, long n, long mod) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = llMod(ans, a, mod);
            }
            a = llMod(a, a, mod);
            n >>= 1;
        }
        return (int) ans;
    }

    private static long[][] initC(int n) {
        long c[][] = new long[n][n];

        for (int i = 0; i < n; i++) {
            c[i][0] = 1;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= i; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
        return c;
    }

    /**
     * ps: n >= m, choose m from n;
     */
    private static int c(long n, long m) {
        if (m > n) {
            n ^= m;
            m ^= n;
            n ^= m;
        }
        m = Math.min(m, n - m);

        long top = 1;
        long bot = 1;
        for (long i = n - m + 1; i <= n; i++) {
            top = (top * i) % MOD;
        }
        for (int i = 1; i <= m; i++) {
            bot = (bot * i) % MOD;
        }

        return (int) ((top * pow(bot, MOD - 2)) % MOD);
    }

    static int gcd(int a, int b) {
        if (a < b) {
            a ^= b;
            b ^= a;
            a ^= b;
        }
        while (b != 0) {
            int tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }

    static boolean even(long n) {
        return (n & 1) == 0;
    }

    public static void main(String args[]) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        st = new StreamTokenizer(br);
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        st.ordinaryChar('\''); //指定单引号、双引号和注释符号是普通字符
        st.ordinaryChar('\"');
        st.ordinaryChar('/');

        solve();
        pw.flush();
    }

    static String next(int len) throws IOException {
        int b = br.read();
        while (b == '\n' || b == ' ' || b == '\r' || b == '\t') {
            b = br.read();
        }
        char ch[] = new char[len];
        ch[0] = (char) b;
        int idx = 1;
        while (idx < len && (b = br.read()) != ' ' && b != '\n' && b != '\r' && b != '\t') {
            ch[idx++] = (char) b;
        }
        return String.valueOf(ch).trim();
    }

    private static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    private static double nextDouble() throws IOException {
        st.nextToken();
        return st.nval;
    }

    private static String[] nextSS(String reg) throws IOException {
        return br.readLine().split(reg);
    }

    private static String nextLine() throws IOException {
        return br.readLine();
    }
}