// Author: Ondra Sladký
// State-based solution in O(w^2h^2).
// It should solve the first two subtasks.
#include <bits/stdc++.h>
using namespace std;
#define f(i, a, b) for (int i = (a); i < (b); i++)
#define v vector
int w, h, k, l;
int xhI, yhI, xvI, yvI;
int xt, yt;
v<v<bool>> grid;
v<v<bool>> validH;
v<v<bool>> validV;
v<v<v<v<bool>>>> states;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> w >> h >> k >> l;
cin >> xhI >> yhI >> xvI >> yvI;
grid = v<v<bool>>(w, v<bool>(h));
f(i, 0, h) {
string s;
cin >> s;
f(j, 0, w) {
char x = s[j];
grid[j][i] = x == 'X';
if (x == '*') {
xt = j;
yt = i;
}
}
}
// Precompute valid positions for horizontal and vertical blocks.
validH = v<v<bool>>(w, v<bool>(h, false));
validV = v<v<bool>>(w, v<bool>(h, false));
f(i,0,h) {
int last = -1;
f(j,0,w) {
if (grid[j][i]) {
last = j;
}
int start = j - k + 1;
if (start > last) {
validH[start][i] = true;
}
}
}
f(i,0,w) {
int last = -1;
f(j,0,h) {
if (grid[i][j]) {
last = j;
}
int start = j - l + 1;
if (start > last) {
validV[i][start] = true;
}
}
}
auto directions = v<tuple<int, int, int>> {
{0, 1, 0},
{0, -1, 0},
{0, 0, 1},
{0, 0, -1},
{1, 1, 0},
{1, -1, 0},
{1, 0, 1},
{1, 0, -1},
};
queue<tuple<int, int, int, int>> q;
q.push({xhI, yhI, xvI, yvI});
states = v<v<v<v<bool>>>>(w, v<v<v<bool>>>(h, v<v<bool>>(w, v<bool>(h, false))));
states[xhI][yhI][xvI][yvI] = true;
while (!q.empty()) {
auto [xh, yh, xv, yv] = q.front();
q.pop();
for (auto [obj, dx, dy] : directions) {
int nxh = xh + obj * dx;
int nyh = yh + obj * dy;
int nxv = xv + (!obj) * dx;
int nyv = yv + (!obj) * dy;
// Fast check for obstacles
bool blocked = false;
if (nxh < 0 || nxh >= w || nyh < 0 || nyh >= h) blocked = true;
if (nxv < 0 || nxv >= w || nyv < 0 || nyv >= h) blocked = true;
if (blocked) continue;
if (!validH[nxh][nyh]) blocked = true;
if (!validV[nxv][nyv]) blocked = true;
// Check if the objects overlap.
int nddx = nxv - nxh;
int nddy = nyh - nyv;
if (nddx < 0 || nddx >= k || nddy < 0 || nddy >= l) blocked = true;
if (blocked) continue;
if (states[nxh][nyh][nxv][nyv]) continue;
states[nxh][nyh][nxv][nyv] = true;
q.push({nxh, nyh, nxv, nyv});
}
}
bool reached = false;
f(i, 0, w) f(j, 0, h) {
if (states[i][yt][xt][j]) {
reached = true;
break;
}
}
cout << (reached ? "YES" : "NO") << endl;
}
