import heapq

# Input
N, V, OY, OX = map(int, input().split())
L = [ list(map(int, input().split())) for i in range(N) ]
OX, OY = OX - 1, OY - 1

# Dijkstra
dx = [ 0, 1, 0, -1 ]
dy = [ 1, 0, -1, 0 ]
dist = [ [ [ 0 ] * N for j in range(N) ] for i in range(2) ]
vis = [ [ [ False ] * N for j in range(N) ] for i in range(2) ]
que = [ (-V, 0, 0, 0) ]
heapq.heapify(que)
dist[0][0][0] = V
while len(que) >= 1:
	u = heapq.heappop(que)
	if vis[u[1]][u[2]][u[3]]:
		continue
	vis[u[1]][u[2]][u[3]] = True
	for i in range(4):
		tx, ty = u[2] + dx[i], u[3] + dy[i]
		if 0 <= tx and tx < N and 0 <= ty and ty < N:
			ncost = -u[0] - L[tx][ty]
			tc = u[1]
			if tc == 0 and tx == OX and ty == OY:
				ncost *= 2
				tc = 1
			if dist[tc][tx][ty] < ncost:
				dist[tc][tx][ty] = ncost
				heapq.heappush(que, (-ncost, tc, tx, ty))

# Output
if dist[0][N - 1][N - 1] > 0 or dist[1][N - 1][N - 1] > 0:
	print("YES")
else:
	print("NO")