#include <bits/stdc++.h>
using namespace std;
int main() {
  stack<int> x, y;
  int m, n;
  string a, b, ans;
  cin >> m >> n;
  int nowa = m - n, nowb = m - n;
  cin >> a >> b;
  ans = a;
  for (int i = 0; i < a.size(); i++) {
    if (a[i] != 'a' && b[i] != 'a') ans[i] = 'a';
    if (a[i] != 'b' && b[i] != 'b') ans[i] = 'b';
    if (a[i] != 'c' && b[i] != 'c') ans[i] = 'c';
    if (a[i] == b[i])
      x.push(i);
    else
      y.push(i);
  }
  while (x.size() && nowa) ans[x.top()] = a[x.top()], nowa--, nowb--, x.pop();
  while (y.size() && nowa) ans[y.top()] = a[y.top()], nowa--, y.pop();
  while (y.size() && nowb) ans[y.top()] = b[y.top()], nowb--, y.pop();
  if (nowb) {
    cout << -1 << endl;
    return 0;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  stack<int> x, y;
  int m, n;
  string a, b, ans;
  cin >> m >> n;
  int nowa = m - n, nowb = m - n;
  cin >> a >> b;
  ans = a;
  for (int i = 0; i < a.size(); i++) {
    if (a[i] != 'a' && b[i] != 'a') ans[i] = 'a';
    if (a[i] != 'b' && b[i] != 'b') ans[i] = 'b';
    if (a[i] != 'c' && b[i] != 'c') ans[i] = 'c';
    if (a[i] == b[i])
      x.push(i);
    else
      y.push(i);
  }
  while (x.size() && nowa) ans[x.top()] = a[x.top()], nowa--, nowb--, x.pop();
  while (y.size() && nowa) ans[y.top()] = a[y.top()], nowa--, y.pop();
  while (y.size() && nowb) ans[y.top()] = b[y.top()], nowb--, y.pop();
  if (nowb) {
    cout << -1 << endl;
    return 0;
  }
  cout << ans << endl;
  return 0;
}

```