n,m = map(int, raw_input().split())

start = [raw_input() for __ in xrange(n)]
finish = [raw_input() for __ in xrange(n)]

count_row = [0 for __ in xrange(n)]
count_col = [0 for __ in xrange(m)]
for i in xrange(n):
	count_row[i] = sum(finish[i][j] == 'X' for j in xrange(m))
for j in xrange(m):
	count_col[j] = sum(finish[i][j] == 'X' for i in xrange(n))

forbidden_rows = set()
forbidden_cols = set()
# if there are two down pegs in a row or column, we can't ever perform move on the rows and cols of those pegs
for i in xrange(n):
	for j in xrange(m):
		if finish[i][j] == 'X' and (count_row[i] >= 2 or count_col[j] >= 2):
			forbidden_rows.add(i)
			forbidden_cols.add(j)


# for each nonforbidden row/col, it can only have at most one down peg that is in a nonforbidden col/row.
haszero = False
hasnonzero = False
same = True
for i in xrange(n):
	for j in xrange(m):
		if i not in forbidden_rows and j not in forbidden_cols:
			haszero = haszero or finish[i][j] == 'X'
			hasnonzero = hasnonzero or start[i][j] == 'O'
			same = same and start[i][j] == finish[i][j]

# so far, we are ok if we are all the same, or there is some down peg in finish not in a forbidden row/col and some up peg in start
ok = same or (haszero and hasnonzero)

# now we look at things that lie in a forbidden row or column and check for differences
for i in xrange(n):
	for j in xrange(m):
		if (i in forbidden_rows or j in forbidden_cols) and start[i][j] != finish[i][j]:
			if finish[i][j] == 0 or (i in forbidden_rows and j in forbidden_cols):
				# if this is hammered down, it is impossible since we can't fix it directly (might not need to check since this implies second condition will be true)
				# if this is in a forbidden row and column, there is no way to fix it indirectly
				ok = False
			else:
				# we might be able to use an allowed row or column to pop this peg back up
				# can only do this if there is up peg in start and down peg in finish
				ok = ok and (haszero and hasnonzero)

print 1 if ok else 0