import sys
from collections import deque

R, C = map(int, sys.stdin.readline().split())
sy, sx = map(int, sys.stdin.readline().split())
gy, gx = map(int, sys.stdin.readline().split())

# Convert to 0-based indices
sy -= 1
sx -= 1
gy -= 1
gx -= 1

maze = []
for _ in range(R):
    row = sys.stdin.readline().strip()
    maze.append(list(row))

# Initialize distance matrix with -1 (unvisited)
dist = [[-1 for _ in range(C)] for _ in range(R)]
dist[sy][sx] = 0

queue = deque()
queue.append((sy, sx))

# Directions: up, down, left, right
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

while queue:
    y, x = queue.popleft()
    current_distance = dist[y][x]
    
    for dy, dx in directions:
        ny = y + dy
        nx = x + dx
        
        # Check if within bounds
        if 0 <= ny < R and 0 <= nx < C:
            # Check if the cell is passable and not visited
            if maze[ny][nx] == '.' and dist[ny][nx] == -1:
                dist[ny][nx] = current_distance + 1
                # Check if reached the goal
                if ny == gy and nx == gx:
                    print(dist[ny][nx])
                    sys.exit()
                queue.append((ny, nx))

# According to the problem statement, the goal is always reachable, so this line is theoretically unreachable
print(-1)