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;
char ch(char f, char s) {
  for (char j = 'a'; j <= 'z'; j++) {
    if (j != f && j != s) {
      return j;
    }
  }
}
int main() {
  int n, t;
  cin >> n >> t;
  string s1, s2, s3 = "";
  cin >> s1 >> s2;
  vector<int> s, d;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      s3 += s1[i];
      s.push_back(i);
    } else {
      s3 += ch(s1[i], s2[i]);
      d.push_back(i);
    }
  }
  int x = d.size();
  for (int i = 0; i < s.size(); i++) {
    if (x >= t) break;
    s3[s[i]] = ch(s1[s[i]], s1[s[i]]);
    ++x;
  }
  for (int i = 0; i + 1 < d.size(); i += 2) {
    if (x <= t) break;
    s3[d[i]] = s1[d[i]];
    s3[d[i + 1]] = s2[d[i + 1]];
    --x;
  }
  if (x != t)
    cout << -1;
  else
    cout << s3;
  return 0;
}