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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

/*
#include <boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
using cint = cpp_int;
*/

// Define
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const ll dx[4] = {1, 0, -1, 0};
const ll dy[4] = {0, 1, 0, -1};
const ll MOD = 1e9 + 7;
const ll mod = 998244353;
const ll inf = 1 << 30;
// const ll INF = LONG_MAX;
const ll INF = 1LL << 60;
const ull MAX = ULONG_MAX;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define elif else if
#define endl '\n'
#define space ' '
#define def inline auto
#define func inline constexpr ll
#define run(a) __attribute__((constructor)) def _##a()
#define all(v) begin(v), end(v)
#define input(a) scanf("%lld", &(a))
#define print(a) printf("%lld\n", (a))
#define fi first
#define se second
#define ok(a, b) (0 <= (a) && (a) < (b))
template <class T> using vvector = vector<vector<T>>;
template <class T> using pvector = vector<pair<T, T>>;
template <class T>
using rpriority_queue = priority_queue<T, vector<T>, greater<T>>;

template <class T> bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> bool chmin(T &a, const T &b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

// Debug
#define debug(...)                                                             \
    {                                                                          \
        cerr << __LINE__ << ": " << #__VA_ARGS__ << " = ";                     \
        for (auto &&X : {__VA_ARGS__}) cerr << "[" << X << "] ";               \
        cerr << endl;                                                          \
    }

#define dump(a, h, w)                                                          \
    {                                                                          \
        cerr << __LINE__ << ": " << #a << " = [" << endl;                      \
        rep(__i, h) {                                                          \
            rep(__j, w) cerr << a[__i][__j] << space;                          \
            cerr << endl;                                                      \
        }                                                                      \
        cerr << "]" << endl;                                                   \
    }

#define vdump(a, n)                                                            \
    {                                                                          \
        cerr << __LINE__ << ": " << #a << " = [";                              \
        rep(__i, n) if (__i) cerr << space << a[__i];                          \
        else cerr << a[__i];                                                   \
        cerr << "]" << endl;                                                   \
    }

struct edge {
    ll to, cost;
    edge(ll a, ll b) : to(a), cost(b) {}
};

struct position {
    ll x, y;
    position() {}
    position(ll a, ll b) : x(a), y(b) {}
    position next(ll i) { return {x + dx[i], y + dy[i]}; }
    ll mdist() { return abs(x) + abs(y); }
    double dist() { return sqrt(x * x + y * y); }
    double norm(ll d) {
        if (d == inf) return max(x, y);
        if (d == 1) return mdist();
        if (d == 2) return dist();
        return 0;
    }
    ll num(ll width) { return abs(x) * width + abs(y); }

    bool operator==(position a) { return x == a.x && y == a.y; }
    bool operator!=(position a) { return x != a.x || y != a.y; }
    bool operator<(position a) { return x < a.x && y < a.y; }
    bool operator>(position a) { return x > a.x && y > a.y; }
    bool operator<=(position a) { return x <= a.x && y <= a.y; }
    bool operator>=(position a) { return x >= a.x && y >= a.y; }
    position operator+(position a) { return position(x + a.x, y + a.y); }
    position operator-(position a) { return position(x - a.x, y - a.y); }
    position operator*(position a) { return position(x * a.x, y * a.y); }
    position operator/(position a) { return position(x / a.x, y / a.y); }
    position operator%(position a) { return position(x % a.x, y % a.y); }
    position complex(position a) {
        return position(x * a.x - y * a.y, x * a.y + y * a.x);
    }
    /*
        // for sort:
        bool operator<(position a) { return x ^ a.x ? x < a.x : y < a.y; }
        bool operator>(position a) { return x ^ a.x ? x > a.x : y > a.y; }
        bool operator<=(position a) { return x ^ a.x ? x < a.x : y <= a.y; }
        bool operator>=(position a) { return x ^ a.x ? x > a.x : y >= a.y; }
    */
};
position Origin = position(0, 0);
using pos = position;
using vec = position;

struct Range {
    ll left, right;
    Range() {}
    Range(ll l, ll r) : left(l), right(r) {}
    ll length() { return right - left; }
    bool operator==(Range A) { return left == A.left && right == A.right; }
    bool operator!=(Range A) { return !(Range(left, right) == A); }
    bool operator>(Range A) { return left < A.left && right > A.right; }
    bool operator<(Range A) { return left > A.left && right < A.right; }
    bool operator>=(Range A) { return left <= A.left && right >= A.right; }
    bool operator<=(Range A) { return left >= A.left && right <= A.right; }
};

// Loop
#define inc(i, a, n) for (ll i = (a), _##i = (n); i <= _##i; ++i)
#define dec(i, a, n) for (ll i = (a), _##i = (n); i >= _##i; --i)
#define rep(i, n) for (ll i = 0, _##i = (n); i < _##i; ++i)
#define each(i, a) for (auto &&i : a)
#define loop() for (;;)

// Stream
#define fout(n) cout << fixed << setprecision(n)
#define fasten cin.tie(0), ios::sync_with_stdio(0)

// Speed
run(0) { fasten, fout(10); }
#pragma GCC optimize("O3")
#pragma GCC target("avx")

// Gen-Test
#define RAND(a)                                                                \
    {                                                                          \
        random_device rnd;                                                     \
        mt19937_64 a(rnd());                                                   \
    }
#define warning(A)                                                             \
    if (!(A)) return 1
#define sin(a, b) ifstream a(b);
#define sout(a, b) ofstream a(b);

// Math
//#define gcd __gcd
func gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
func lcm(ll a, ll b) { return a * b / gcd(a, b); }
func sign(ll a) { return a ? abs(a) / a : 0; }

template <class T> def in() {
    T A;
    cin >> A;
    return A;
}

template <class T>
def out(vector<vector<T>> A, ll H, ll W, char divc = space, char endc = endl) {
    rep(i, H) {
        rep(j, W) {
            if (j)
                cout << divc << A[i][j];
            else
                cout << A[i][j];
        }
        cout << endc;
    }
}

bool check(vector<ll> A) {
    return A[0] <= A[1] && A[2] <= A[3];
}

signed main() {
    ll H, W, N, X, Y, res = 0;
    string S1, S2;
    cin >> H >> W >> N >> X >> Y >> S1 >> S2;
    vector<ll> A = {0, H - 1, 0, W - 1};
    dec(i, N - 1, 0) {
        if (S2[i] == 'L') {
            if (A[2] > 0) A[2]--;
        } else if (S2[i] == 'R') {
            if (A[3] < W - 1) A[3]++;
        } else if (S2[i] == 'U') {
            if (A[0] > 0) A[0]--;
        } else if (S2[i] == 'D') {
            if (A[1] < H - 1) A[1]++;
        }
        if (!check(A)) res = 1;
        if (S1[i] == 'L') {
            A[3]--;
        } else if (S1[i] == 'R') {
            A[2]++;
        } else if (S1[i] == 'U') {
            A[1]--;
        } else if (S1[i] == 'D') {
            A[0]++;
        }
        if (!check(A)) res = 1;
    }
    if (A[0] > X - 1 || A[1] < X - 1 || A[2] > Y - 1 || A[3] < Y - 1) res = 1;
    cout << (!res ? "YES" : "NO") << endl;
}

// for compilation: g++ -Ofast -march=native -o _ _.cpp -std=c++17