#include <bits/stdc++.h>
using namespace std;
char change(char a, char b) {
  for (char i = 'a'; i <= 'z'; ++i)
    if (i != a && i != b) return i;
}
int main() {
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  vector<int> a, b;
  int d = 0;
  for (int i = 0; i < n; ++i)
    if (s1[i] == s2[i])
      a.push_back(i);
    else {
      b.push_back(i);
      d++;
    }
  if (d > 2 * t) {
    cout << -1 << endl;
    return 0;
  }
  if (t < d && d <= 2 * t) {
    for (int i = 0; i < 2 * (d - t); i += 2) s1[b[i]] = s2[b[i]];
    for (int i = 2 * (d - t); i < d; ++i)
      s1[b[i]] = s2[b[i]] = change(s1[b[i]], s2[b[i]]);
    cout << s1 << endl;
    return 0;
  }
  for (int i = 0; i < d; ++i) s1[b[i]] = s2[b[i]] = change(s1[b[i]], s2[b[i]]);
  for (int i = 0; i < t - d; ++i)
    s1[a[i]] = s2[a[i]] = change(s1[a[i]], s2[a[i]]);
  cout << s1 << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
char change(char a, char b) {
  for (char i = 'a'; i <= 'z'; ++i)
    if (i != a && i != b) return i;
}
int main() {
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  vector<int> a, b;
  int d = 0;
  for (int i = 0; i < n; ++i)
    if (s1[i] == s2[i])
      a.push_back(i);
    else {
      b.push_back(i);
      d++;
    }
  if (d > 2 * t) {
    cout << -1 << endl;
    return 0;
  }
  if (t < d && d <= 2 * t) {
    for (int i = 0; i < 2 * (d - t); i += 2) s1[b[i]] = s2[b[i]];
    for (int i = 2 * (d - t); i < d; ++i)
      s1[b[i]] = s2[b[i]] = change(s1[b[i]], s2[b[i]]);
    cout << s1 << endl;
    return 0;
  }
  for (int i = 0; i < d; ++i) s1[b[i]] = s2[b[i]] = change(s1[b[i]], s2[b[i]]);
  for (int i = 0; i < t - d; ++i)
    s1[a[i]] = s2[a[i]] = change(s1[a[i]], s2[a[i]]);
  cout << s1 << endl;
  return 0;
}

```