#include <bits/stdc++.h>
using namespace std;
const long long OO = 1e9;
char s3[100005];
int main() {
  string s1, s2;
  int n, t;
  cin >> n >> t;
  cin >> s1 >> s2;
  int k1 = n - t, k2 = n - t, t1 = t, t2 = t;
  for (int i = 0; i < n && k1 && k2; i++)
    if (s1[i] == s2[i]) {
      k1--, k2--;
      s3[i] = s1[i];
    }
  for (int i = 0; i < n; i++) {
    if (s3[i] == 0) {
      if (k1)
        s3[i] = s1[i], k1--, t1--;
      else if (k2)
        s3[i] = s2[i], k2--, t2--;
      else {
        t1--, t2--;
        char x = 'a';
        while (!(x != s1[i] && x != s2[i])) x++;
        s3[i] = x;
      }
    }
  }
  if (!t1 && !t2 && !k1 && !k2)
    for (int i = 0; i < n; i++) cout << s3[i];
  else
    cout << -1;
}

### Prompt

Please formulate 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 long long OO = 1e9;
char s3[100005];
int main() {
  string s1, s2;
  int n, t;
  cin >> n >> t;
  cin >> s1 >> s2;
  int k1 = n - t, k2 = n - t, t1 = t, t2 = t;
  for (int i = 0; i < n && k1 && k2; i++)
    if (s1[i] == s2[i]) {
      k1--, k2--;
      s3[i] = s1[i];
    }
  for (int i = 0; i < n; i++) {
    if (s3[i] == 0) {
      if (k1)
        s3[i] = s1[i], k1--, t1--;
      else if (k2)
        s3[i] = s2[i], k2--, t2--;
      else {
        t1--, t2--;
        char x = 'a';
        while (!(x != s1[i] && x != s2[i])) x++;
        s3[i] = x;
      }
    }
  }
  if (!t1 && !t2 && !k1 && !k2)
    for (int i = 0; i < n; i++) cout << s3[i];
  else
    cout << -1;
}

```