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:
#!/usr/bin/python3

n, t = map(int, input().split())
a, b = input(), input()
k = len([1 for i, j in zip(a, b) if i != j])
if t < (k + 1) // 2:
    print(-1)
else:
    p = -1
    ans = ['0'] * n
    for i in range(n):
        if a[i] != b[i] and k > t:
            if p == -1:
                p = i
            else:
                ans[p] = a[p]
                ans[i] = b[i]
                p = -1
                k -= 2
                t -= 1
    for i in range(n):
        if a[i] != b[i] and ans[i] == '0':
            for v in 'abc':
                if v != a[i] and v != b[i]:
                    ans[i] = v
                    break
            t -= 1
    for i in range(n):
        if a[i] == b[i] and t:
            ans[i] = 'a' if 'a' != a[i] else 'b'
            t -= 1
        elif a[i] == b[i]:
            ans[i] = a[i]
    print(''.join(ans))