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/env python3
# -*- coding: utf-8 -*-
#C
n, t = map(int,input().split())
s1 = list(input())
s2 = list(input())
s = list(s1)

#s1とs2の違う文字数
dif_1_2 = 0
same_1_2 = 0
dif_place = []
same_place = []
for i in range(n):
    if s1[i] != s2[i]:
        dif_1_2 += 1
        dif_place.append(i)
    else:
        same_place.append(i)

same_ans_1 = len(s)
#print(dif_1_2)

if dif_1_2 == t:
    for i in dif_place:
        if s1[i] != 'a' and s2[i] != 'a':
            s[i] = 'a'
        elif s1[i] != 'b' and s2[i] != 'b':
            s[i] = 'b'
        elif s1[i] != 'c' and s2[i] != 'c':
            s[i] = 'c'

    print("".join(s))
    exit()

if dif_1_2 > t:
    if dif_1_2 > 2*t:
        print(-1)
        exit()

    change_cnt = 0

    for i in dif_place:
        if dif_1_2 > t:
            if change_cnt % 2 ==0:
                s[i] = s1[i]
            else:
                s[i] = s2[i]
                dif_1_2 -= 1
            change_cnt += 1

        else:
            if s1[i] != 'a' and s2[i] != 'a':
                s[i] = 'a'
            elif s1[i] != 'b' and s2[i] != 'b':
                s[i] = 'b'
            elif s1[i] != 'c' and s2[i] != 'c':
                s[i] = 'c'

    print("".join(s))
    exit()

dif = dif_1_2
if dif_1_2 < t:
    for i in dif_place:
        if s1[i] != 'a' and s2[i] != 'a':
            s[i] = 'a'
        elif s1[i] != 'b' and s2[i] != 'b':
            s[i] = 'b'
        elif s1[i] != 'c' and s2[i] != 'c':
            s[i] = 'c'

    for i in same_place:
        if dif == t:
            break
        if s1[i] != 'a':
            s[i] = 'a'
        else:
            s[i] = 'b'
        dif += 1
    print("".join(s))
    exit()