#include <bits/stdc++.h>
using namespace std;
int n, t, k, q;
char s1[100001], s2[100001], s3[100001];
char fix(char a, char b) {
  char i;
  for (i = 'a'; i <= 'z'; i++)
    if (i != a && i != b) return i;
  return i;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> t;
  cin >> s1 >> s2;
  for (int i = 0; i < n; i++) s3[i] = fix(s1[i], s2[i]);
  int ds1 = n, ds2 = n;
  for (int i = 0; i < n; i++) {
    if (ds1 == t) break;
    if (s1[i] == s2[i]) {
      s3[i] = s1[i];
      ds1--;
      ds2--;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      if (ds1 > t) {
        ds1--;
        s3[i] = s1[i];
      } else if (ds2 > t) {
        ds2--;
        s3[i] = s2[i];
      }
    }
  }
  if (ds1 != t || ds2 != t) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; i++) cout << s3[i];
}

### Prompt

Your challenge is to write 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;
int n, t, k, q;
char s1[100001], s2[100001], s3[100001];
char fix(char a, char b) {
  char i;
  for (i = 'a'; i <= 'z'; i++)
    if (i != a && i != b) return i;
  return i;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> t;
  cin >> s1 >> s2;
  for (int i = 0; i < n; i++) s3[i] = fix(s1[i], s2[i]);
  int ds1 = n, ds2 = n;
  for (int i = 0; i < n; i++) {
    if (ds1 == t) break;
    if (s1[i] == s2[i]) {
      s3[i] = s1[i];
      ds1--;
      ds2--;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      if (ds1 > t) {
        ds1--;
        s3[i] = s1[i];
      } else if (ds2 > t) {
        ds2--;
        s3[i] = s2[i];
      }
    }
  }
  if (ds1 != t || ds2 != t) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; i++) cout << s3[i];
}

```