#include <deque>
#include <iostream>
#include <vector>
using namespace std;
int main() {
    int h, w;
    cin >> h >> w;
    vector v(h + 2, vector<char>(w + 2, '#'));
    long u, d, r, l, k, p;
    cin >> u >> d >> r >> l >> k >> p;
    int xs, ys, xt, yt;
    cin >> xs >> ys >> xt >> yt;
    for (int i = 1; i < h + 1; i++) {
        for (int j = 1; j < w + 1; j++) {
            cin >> v[i][j];
        }
    }
    deque<vector<int>> dq;
    v[xt][yt] = 'g';
    dq.push_back({xs, ys, 0});
    vector<int> D[4] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
    long cost = 0;
    bool fg = 0;
    while (!dq.empty()) {
        auto fr = dq.front();
        dq.pop_front();
        v[fr[0]][fr[1]] = '*';
        if (fr[2] == 1) {
            cost += p;
        }
        for (int i = 0; i < 4; i++) {
            if (v[fr[0] + D[i][0]][fr[1] + D[i][1]] == 'g') {
                fg = 1;
                break;
            } else if (v[fr[0] + D[i][0]][fr[1] + D[i][1]] == '.') {
                dq.push_front({fr[0] + D[i][0], fr[1] + D[i][1], 0});
            } else if (v[fr[0] + D[i][0]][fr[1] + D[i][1]] == '@') {
                dq.push_back({fr[0] + D[i][0], fr[1] + D[i][1], 1});
            }
        }
    }
    if (!fg) {
        cout << "No" << endl;
        return 0;
    }
    dq.clear();
    dq.push_back({xs, ys});
    while (!dq.empty()) {
        auto fr = dq.front();
        dq.pop_front();
        v[fr[0]][fr[1]] = '+';
        for (int i = 0; i < 4; i++) {
            if (v[fr[0] + D[i][0]][fr[1] + D[i][1]] == 'g') {
                if (i == 0) {
                    cost += u;
                } else if (i == 1) {
                    cost += d;
                } else if (i == 2) {
                    cost += r;
                } else {
                    cost += l;
                }
                break;
            } else if (v[fr[0] + D[i][0]][fr[1] + D[i][1]] == '*') {
                if (i == 0) {
                    cost += u;
                } else if (i == 1) {
                    cost += d;
                } else if (i == 2) {
                    cost += r;
                } else {
                    cost += l;
                }
                dq.push_back({fr[0] + D[i][0], fr[1] + D[i][1]});
            }
        }
    }
    if (cost <= k) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
    return 0;
}