Problem: Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.

More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print  - 1.

Input

The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n).

The second line contains string s1 of length n, consisting of lowercase English letters.

The third line contain string s2 of length n, consisting of lowercase English letters.

Output

Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.

Examples

Input

3 2
abc
xyc


Output

ayd

Input

1 0
c
b


Output

-1

Solution (in PYTHON3):

Solution:
n, t = map(int, input().split())
s1 = input()
s2 = input()
sm = n-t

check = []
ans = ['#']*n
same = 0
diff = 0
for i in range(n):
	if s1[i] == s2[i]:
		check.append(True)
		same += 1
	else:
		check.append(False)
		diff += 1

if same + diff//2 < sm:
	print(-1)
else:
	i = 0
	t1 = t2 = sm
	while i<n and t1 and t2:
		if check[i]:
			ans[i] = s1[i]
			t1 -= 1
			t2 -= 1
		i += 1

	i = 0			
	while i<n and t1:
		if ans[i] == '#':
			ans[i] = s1[i]
			t1 -= 1
		i += 1

	i = 0			
	while i<n and t2:
		if ans[i] == '#':
			ans[i] = s2[i]
			t2 -= 1
		i += 1

	for i in range(n):
		if ans[i] == '#':
			if s1[i]!='a' and s2[i]!='a':
				ans[i] = 'a'
			elif s1[i]!='b' and s2[i]!='b':
				ans[i] = 'b'
			else:
				ans[i] = 'c'

	t1 = t2 = 0
	for i in range(n):
		if ans[i] != s1[i]:
			t1 += 1
		if ans[i] != s2[i]:
			t2 += 1

	if t == t1 == t2:
		print(''.join(ans))
	else:
		print(-1)