#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
using namespace std;
using ll = long long;
using vec = vector<ll>;
using vect = vector<double>;
using Graph = vector<vector<ll>>;
#define loop(i, n) for (ll i = 0; i < n; i++)
#define Loop(i, m, n) for (ll i = m; i < n; i++)
#define pool(i, n) for (ll i = n; i >= 0; i--)
#define Pool(i, m, n) for (ll i = n; i >= m; i--)
#define mod 1000000007ll
//#define mod 998244353ll
#define flagcount __builtin_popcount
#define flag(x) (1 << x)
#define flagadd(bit, x) bit |= flag(x)
#define flagpop(bit, x) bit &= ~flag(x)
#define flagon(bit, i) bit &flag(i)
#define flagoff(bit, i) !(bit & (1 << i))
#define all(v) v.begin(), v.end()
#define low2way(v, x) lower_bound(all(v), x)
#define high2way(v, x) upper_bound(all(v), x)
#define idx_lower(v, x) (distance(v.begin(), low2way(v, x)))  //配列vでx未満の要素数を返す
#define idx_upper(v, x) (distance(v.begin(), high2way(v, x))) //配列vでx以下の要素数を返す
#define idx_lower2(v, x) (v.size() - idx_lower(v, x))         //配列vでx以上の要素数を返す
#define idx_upper2(v, x) (v.size() - idx_upper(v, x))         //配列vでxより大きい要素の数を返す
#define putout(a) cout << a << endl
#define Gput(a, b) G[a].push_back(b)
#define Sum(v) accumulate(all(v), 0ll)
#define gcd(x, y) __gcd(x, y)
ll ctoi(char c)
{
    if (c >= '0' && c <= '9')
    {
        return c - '0';
    }
    return 0;
}
template <typename T>
T lcm(T x, T y)
{
    T z = gcd(x, y);
    return x * y / z;
}
template <typename T>
bool primejudge(T n)
{
    if (n < 2)
        return false;
    else if (n == 2)
        return true;
    else if (n % 2 == 0)
        return false;
    double sqrtn = sqrt(n);
    for (T i = 3; i < sqrtn + 1; i++)
    {
        if (n % i == 0)
        {
            return false;
        }
        i++;
    }
    return true;
}
//場合によって使い分ける
//const ll dx[4]={1,0,-1,0};
//const ll dy[4]={0,1,0,-1};
const ll dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
//2次元配列の宣言
//vector<vector<ll>> field(h, vector<ll>(w));

int main()
{
    cout << fixed << setprecision(30);
    ll H, W, N, sx, sy;
    cin >> H >> W >> N >> sx >> sy;
    sx--;
    sy--;
    string S, T;
    cin >> S >> T;
    //駒を盤面に残すことができるかを考える
    bool ans = true;
    //左右の移動のみを考える(区間は閉区間[l,r]で考える)
    ll l = 0, r = W - 1;
    //T[N-1]は結果に影響しない
    if (S[N - 1] == 'L')
        l++;
    if (S[N - 1] == 'R')
        r--;
    pool(i, N - 2)
    {
        if (T[i] == 'L')
            r = min(r + 1, W - 1);
        if (T[i] == 'R')
            l = max(l - 1, 0ll);
        if (S[i] == 'L')
            l++;
        if (S[i] == 'R')
            r--;
        if (l > r)
            ans = false;
    }
    if (sy < l || sy > r)
        ans = false;
    //今度は上下の移動のみを考える
    ll L = 0, R = H - 1;
    if (S[N - 1] == 'U')
        L++;
    if (S[N - 1] == 'D')
        R--;
    pool(i, N - 2)
    {
        if (T[i] == 'U')
            R = min(R + 1, H - 1);
        if (T[i] == 'D')
            L = max(L - 1, 0ll);
        if (S[i] == 'U')
            L++;
        if (S[i] == 'D')
            R--;
        if (L > R)
            ans = false;
    }
    if (sx < L || sx > R)
        ans = false;
    if (ans)
        putout("YES");
    else
        putout("NO");
    return 0;
}


Measured memory_bytes: 3924