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;
string s1, s2, s;
int n, t, m, ans, res;
int main() {
  cin >> n >> t;
  cin >> s1 >> s2;
  for (int i = 0; i < n; i++)
    if (s1[i] == s2[i])
      ans++;
    else
      res++;
  t = n - t;
  m = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      s += s1[i];
      continue;
    }
    if (t <= ans) {
      if (s1[i] != 'a' && s2[i] != 'a')
        s += 'a';
      else if (s1[i] != 'b' && s2[i] != 'b')
        s += 'b';
      else if (s1[i] != 'c' && s2[i] != 'c')
        s += 'c';
    } else {
      if (!m) s += s1[i];
      if (m) s += s2[i];
      m ^= 1;
      if (!m) t--;
    }
  }
  if (m % 2 || t > ans) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (s1[i] == s2[i] && t < ans) {
      s[i] = ('a' + (s[i] - 'a' + 1) % 26);
      t++;
    }
  cout << s << endl;
}