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 get_char(const int a, const int b) {
  for (char c = 'a'; c <= 'z'; ++c) {
    if (c != a && c != b) return c;
  }
  assert(1 > 2);
  return 'a';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  int diff = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) ++diff;
  }
  if (diff > t + t) {
    cout << -1 << '\n';
    return 0;
  }
  string ans(a);
  for (int i = 0, cnt = t; i < n && cnt > 0; ++i) {
    if (a[i] != b[i]) ans[i] = b[i], --cnt;
  }
  for (int i = n - 1, cnt = t; i >= 0 && cnt > 0; --i) {
    if (a[i] != b[i]) {
      if (ans[i] == a[i])
        ans[i] = a[i], --cnt;
      else
        ans[i] = get_char(a[i], b[i]), --cnt;
    }
  }
  for (int i = 0, cnt = t - diff; i < n && cnt > 0; ++i) {
    if (a[i] == b[i]) ans[i] = get_char(a[i], b[i]), --cnt;
  }
  cout << ans << '\n';
  return 0;
}