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;
vector<int> pe;
string s1, s2, s;
int n, k, q1, q2;
int main() {
  cin >> n >> k >> s1 >> s2;
  s.resize(n, '#');
  if (k == 0) {
    if (s1 == s2)
      cout << s1;
    else
      cout << -1;
    return 0;
  }
  k = (n - k);
  for (int i = 0; (i < n) && (k > 0); i++)
    if (s1[i] == s2[i]) {
      s[i] = s1[i];
      k--;
    }
  for (int i = 0; (i < n) && (k > 0); i++) {
    if (s1[i] != s2[i]) {
      s[i] = s1[i];
      i++;
      for (; (i < n); i++)
        if (s1[i] != s2[i]) break;
      if (i == n) {
        cout << -1;
        return 0;
      }
      s[i] = s2[i];
      k--;
    }
  }
  if (k > 0) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (s[i] == '#')
      for (int j = 'a'; j <= 'z'; j++)
        if (j != s1[i] && j != s2[i]) {
          s[i] = j;
          break;
        }
  cout << s;
}