#include <bits/stdc++.h>
using namespace std;
int n, t;
string a;
string b;
string c;
int main() {
  cin >> n >> t;
  cin >> a >> b;
  int u = n;
  c = a;
  for (int i = 0; i < n; i++) {
    char x = 'a';
    if (a[i] == x || b[i] == x) x = 'b';
    if (a[i] == x || b[i] == x) x = 'c';
    c[i] = x;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && t != u) {
      c[i] = a[i];
      u--;
    }
  }
  int z = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] || u == t) continue;
    if (z == 0) {
      c[i] = a[i];
      z = 1;
    } else {
      c[i] = b[i];
      u--;
      z = 0;
    }
  }
  if (u == t)
    cout << c << "\n";
  else
    cout << "-1\n";
}

### Prompt

Create a solution in cpp for 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;
string a;
string b;
string c;
int main() {
  cin >> n >> t;
  cin >> a >> b;
  int u = n;
  c = a;
  for (int i = 0; i < n; i++) {
    char x = 'a';
    if (a[i] == x || b[i] == x) x = 'b';
    if (a[i] == x || b[i] == x) x = 'c';
    c[i] = x;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && t != u) {
      c[i] = a[i];
      u--;
    }
  }
  int z = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] || u == t) continue;
    if (z == 0) {
      c[i] = a[i];
      z = 1;
    } else {
      c[i] = b[i];
      u--;
      z = 0;
    }
  }
  if (u == t)
    cout << c << "\n";
  else
    cout << "-1\n";
}

```