#include <bits/stdc++.h>

using namespace std;

struct Node {
    int at;
    int cost; 
    int prev;
    Node(int at, int cost, int prev) : at(at), cost(cost), prev(prev) {}
    bool operator>(const Node &s) const {
        if (cost != s.cost) return cost > s.cost;
        if (prev != s.prev) return prev > s.prev; //最短経路を辞書順最小にする(省略可)
        return at > s.at;
    }
};

struct Edge {
  int to;
  int cost;  
  Edge(int to, int cost) : to(to), cost(cost) {}  
};

typedef vector<vector<Edge> > AdjList; //隣接リスト

const int INF = 1000000000;
const int NONE = -1;

AdjList graph;

//sは始点、mincは最短経路のコスト、Prevは最短経路をたどる際の前の頂点
void dijkstra(int s, vector<int> &minc, vector<int> &Prev){ 
    priority_queue<Node, vector<Node>, greater<Node> > pq;
    pq.push(Node(s, 0, NONE));
    while(!pq.empty()) {
        Node cur = pq.top();
        pq.pop();
        if (minc[cur.at] <= cur.cost) continue;
        minc[cur.at] = cur.cost;
        Prev[cur.at] = cur.prev;
        for(int i = 0; i < (int)graph[cur.at].size(); i++) {
        	Edge e = graph[cur.at][i];
            if (minc[e.to] > cur.cost + e.cost) {
            	pq.push(Node(e.to, cur.cost + e.cost, cur.at));
            }
        }
    }
}

const int dh[] = {1, -1, 0, 0};
const int dw[] = {0, 0, 1, -1};

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, v, ox, oy;
	cin >> n >> v >> ox >> oy;
	ox--;
	oy--;
	vector< vector<int> > L(n, vector<int>(n));
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			cin >> L[i][j];
		}
	}
	graph.resize(n * n);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			for (int k = 0; k < 4; k++) {
				int h = i + dh[k], w = j + dw[k];
				if (h < 0 || h >= n || w < 0 || w >= n) continue;
				graph[i * n + j].emplace_back(h * n + w, L[h][w]);
			}
		}
	}
	vector<int> minc(n * n, INF), Prev(n * n, NONE);
	dijkstra(0, minc, Prev);
	if (minc[n * n - 1] < v) {
		cout << "YES" << endl;
		return 0;
	}
	if (ox == -1 && oy == -1) {
		cout << "NO" << endl;
		return 0;
	}
	if (minc[oy * n + ox] >= v) {
		cout << "NO" << endl;
		return 0;		
	}
	v -= minc[oy * n + ox];
	v *= 2;
	minc.resize(n * n, INF);
	Prev.resize(n * n, NONE);
	dijkstra(oy * n + ox, minc, Prev);
	cout << (minc[n * n - 1] < v ? "YES" : "NO") << endl;
	return 0;
}