#include <bits/stdc++.h>
using namespace std;
const int OO = 0x3f3f3f3f;
const double eps = (1e-10);
int n, t;
string s1, s2, ans;
int main() {
  ios::sync_with_stdio(false);
  cout.precision(3);
  cin >> n >> t >> s1 >> s2;
  int matches = 0;
  for (int i = 0; i < (int)(n); ++i) matches += (s1[i] == s2[i]);
  if (2 * t < n - matches) {
    cout << -1 << '\n';
    return 0;
  }
  for (int i = 0; i < (int)(n); ++i) {
    ans += 'a';
    if (ans[i] == s1[i] || ans[i] == s2[i]) ++ans[i];
    if (ans[i] == s1[i] || ans[i] == s2[i]) ++ans[i];
  }
  int need1 = n - t, need2 = need1;
  for (int i = 0; i < (int)(n); ++i) {
    if (!(need1 + need2)) break;
    if (s1[i] == s2[i]) ans[i] = s1[i], --need1, --need2;
  }
  if (!(need1 + need2)) {
    cout << ans << '\n';
    return 0;
  }
  for (int i = 0; i < (int)(n); ++i) {
    if (!(need1 + need2)) break;
    if (s1[i] != s2[i]) {
      if (need1)
        ans[i] = s1[i], --need1;
      else
        ans[i] = s2[i], --need2;
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int OO = 0x3f3f3f3f;
const double eps = (1e-10);
int n, t;
string s1, s2, ans;
int main() {
  ios::sync_with_stdio(false);
  cout.precision(3);
  cin >> n >> t >> s1 >> s2;
  int matches = 0;
  for (int i = 0; i < (int)(n); ++i) matches += (s1[i] == s2[i]);
  if (2 * t < n - matches) {
    cout << -1 << '\n';
    return 0;
  }
  for (int i = 0; i < (int)(n); ++i) {
    ans += 'a';
    if (ans[i] == s1[i] || ans[i] == s2[i]) ++ans[i];
    if (ans[i] == s1[i] || ans[i] == s2[i]) ++ans[i];
  }
  int need1 = n - t, need2 = need1;
  for (int i = 0; i < (int)(n); ++i) {
    if (!(need1 + need2)) break;
    if (s1[i] == s2[i]) ans[i] = s1[i], --need1, --need2;
  }
  if (!(need1 + need2)) {
    cout << ans << '\n';
    return 0;
  }
  for (int i = 0; i < (int)(n); ++i) {
    if (!(need1 + need2)) break;
    if (s1[i] != s2[i]) {
      if (need1)
        ans[i] = s1[i], --need1;
      else
        ans[i] = s2[i], --need2;
    }
  }
  cout << ans << '\n';
  return 0;
}

```