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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mul(long long a, long long b) { return (a * b) % (100000007); }
long long add(long long a, long long b) { return (a + b) % (100000007); }
long long sub(long long a, long long b) {
  return (a - b + llabs(a - b) / (100000007) * (100000007) + (100000007)) %
         (100000007);
}
void upd(long long &a, long long b) {
  a = (a % (100000007) + b % (100000007)) % (100000007);
}
char s1[(100000 + 10)], s2[(100000 + 10)];
int n, t, df;
char get(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b') return 'b';
  if (a != 'c' && b != 'c') return 'c';
}
int main() {
  cin >> n >> t;
  scanf("%s%s", s1 + 1, s2 + 1);
  df = 0;
  for (int i = 1; i <= n; i++)
    if (s1[i] ^ s2[i]) ++df;
  if (df > 2 * t) {
    cout << "-1\n";
    return 0;
  }
  if (df <= t) {
    for (int i = 1; i <= n; i++) {
      if (s1[i] != s2[i])
        putchar(get(s1[i], s2[i]));
      else if (t > df)
        putchar(get(s1[i], s2[i])), t--;
      else
        putchar(s1[i]);
    }
  } else {
    int s = (df - t) * 2;
    for (int i = 1; i <= n; i++) {
      if (s1[i] == s2[i])
        putchar(s1[i]);
      else if (!s) {
        putchar(get(s1[i], s2[i]));
      } else if (s & 1)
        putchar(s1[i]), s--;
      else
        putchar(s2[i]), s--;
    }
  }
  return 0;
}