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;
int getDiff(string &s1, string &s2) {
  int res = 0;
  for (int i = 0; i < s1.size(); ++i)
    if (s1[i] != s2[i]) ++res;
  return res;
}
string solve(string &s1, string &s2, int t) {
  int diff1 = 0, diff2 = getDiff(s1, s2);
  string res = s1;
  for (int i = 0; i < s1.size(); ++i) {
    if (diff2 >= t) break;
    if (s1[i] == s2[i]) {
      ++diff1, ++diff2;
      if (s1[i] != 'a' and s2[i] != 'a')
        res[i] = 'a';
      else if (s1[i] != 'b' and s2[i] != 'b')
        res[i] = 'b';
      else
        res[i] = 'c';
    }
  }
  for (int i = 0; i < s1.size(); ++i) {
    if (diff2 <= t) break;
    if (s1[i] != s2[i]) {
      ++diff1, --diff2;
      res[i] = s2[i];
    }
  }
  for (int i = 0; i < s1.size(); ++i) {
    if (diff1 == t) break;
    if (s1[i] != s2[i] and res[i] == s1[i]) {
      ++diff1;
      if (s1[i] != 'a' and s2[i] != 'a')
        res[i] = 'a';
      else if (s1[i] != 'b' and s2[i] != 'b')
        res[i] = 'b';
      else
        res[i] = 'c';
    }
  }
  if (diff1 != t or diff2 != t)
    return "-1";
  else
    return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  cout << solve(s1, s2, t) << '\n';
  return 0;
}