// Original C++ solution (slower):
#include "bits/stdc++.h"

using namespace std;

#ifndef LOCAL

#define endl '\n'

#endif



#define fr(i, a, b) for(int i = a; i <= b; i++)

#define rf(i, a, b) for(int i = a; i >= b; i--)

#define pf push_front

#define pb push_back

#define eb emplace_back

#define fi first

#define se second

#define all(x) x.begin(), x.end()

#define rall(x) x.rbegin(), x.rend()

#define sz(x) (int)x.size()

#define lb lower_bound

#define ub upper_bound



typedef long long ll;

typedef long double f80;

typedef pair<int,int> pii;

typedef pair<ll,ll> pll;



int pct(int x) { return __builtin_popcount(x); }

int pct(ll x) { return __builtin_popcountll(x); }

int bt(int x) { return 31 - __builtin_clz(x); } // floor(log2(x))

int bt(ll x) { return 63 - __builtin_clzll(x); } // floor(log2(x))

int cdiv(int a, int b) { return a / b + !(a < 0 || a % b == 0); }

ll cdiv(ll a, ll b) { return a / b + !(a < 0 || a % b == 0); }

int nxt_C(int x) { int c = x & -x, r = x + c; return (((r ^ x) >> 2) / c) | r; }

ll nxt_C(ll x) { ll c = x & -x, r = x + c; return (((r ^ x) >> 2) / c) | r; }



vector<int> get_bits(int mask) {

	vector<int> bb;

	while(mask) { int b = bt(mask); bb.pb(b); mask ^= (1 << b); }

	reverse(all(bb));

	return bb;

}



int get_mask(vector<int> v) {

	int mask = 0;

	for(int x : v) { mask ^= (1 << x); }

	return mask;

}



template<typename T>

void uniq(vector<T> &v) { sort(all(v)); v.resize(unique(all(v)) - v.begin()); }



mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());



ll rand(ll l, ll r){

	uniform_int_distribution<ll> uid(l, r);

	return uid(rng);

}



void sc() {}



template <typename Head, typename... Tail>

void sc(Head &H, Tail &... T) { cin >> H; sc(T...); }



#ifdef LOCAL

#define debug(...) cerr << "[L:" << __LINE__ << "][" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

#else

#define debug(...) 42

#endif



// #ifndef LOCAL

// string to_string(__int128 x) {

// 	string s = "";

// 	bool neg = 0;

// 	if(x < 0) { s += "-"; neg = 1; x = -x; }

// 	if(!x) s += '0';

// 	while(x) {

// 		int rem = x % 10;

// 		s += to_string(rem);

// 		x /= 10;

// 	}

// 	reverse(s.begin() + neg, s.end());

// 	return s;

// }

// #endif



const int mod = 1e9 + 7; // 998244353;



int pwr(int a,ll b) {

	int ans = 1;

	while(b) {

		if(b & 1) ans = (ans * 1LL * a) % mod;

		a = (a * 1LL * a) % mod;

		b >>= 1;

	}

	return ans;

}



/*

	Lookout for overflows!!

	Check array sizes!!

	Clear before test cases!!

	Use the correct modulo!!

	Check for corner cases!!

	Are you forgetting something?!

	Read problem statement carefully!!!

*/



string s, t;



bool go(char c1, char c2, int d1, int d2, int x, int n) {

    fr(i, 0, sz(s) - 1) {

        if(s[i] == c1) x += d1;

        if(d1 == 1 && x > n) return 1;

        if(d1 == -1 && x < 1) return 1;

        if(t[i] == c2 && x + d2 >= 1 && x + d2 <= n) x += d2;

    }

    return 0;

}



void solve() {

    int h, w, n;

    sc(h, w, n);

    int x, y;

    sc(x, y);

    sc(s, t);

    if(go('L', 'R', -1, 1, y, w))

        { cout << "NO"; return; }

    if(go('R', 'L', 1, -1, y, w))

        { cout << "NO"; return; }

    if(go('U', 'D', -1, 1, x, h))

        { cout << "NO"; return; }

    if(go('D', 'U', 1, -1, x, h))

        { cout << "NO"; return; }

    cout << "YES";

}



int main() {

	ios :: sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);

	int t = 1;

	// cin >> t;

	for(int tt = 1; tt <= t; tt++) {

		solve();

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<algorithm>

using namespace std;



char s1[200001],s2[200001];

int main(){

	int n,m,k;

	int x,y;

	scanf("%d%d%d",&n,&m,&k);

	scanf("%d%d%s%s",&x,&y,s1+1,s2+1);

	int l1=1,r1=n,l2=1,r2=m;

	for(int i=k;i;i--){

		if(s2[i]=='U')r1=min(n,r1+1);

		if(s2[i]=='D')l1=max(1,l1-1);

		if(s2[i]=='L')r2=min(m,r2+1);

		if(s2[i]=='R')l2=max(1,l2-1);

		if(s1[i]=='U')l1++;

		if(s1[i]=='D')r1--;

		if(s1[i]=='L')l2++;

		if(s1[i]=='R')r2--;

		//printf("%d %d %d %d\n",l1,r1,l2,r2);

		if(l1>r1||l2>r2){

			printf("NO\n");

			return 0;

		}

	}

	if(x>=l1&&x<=r1&&y>=l2&&y<=r2)

		printf("YES\n");

	else printf("NO\n");

}

// Fastest known solution for this problem: