#include <bits/stdc++.h>
using namespace std;
char getDist(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 a, b;
  cin >> a >> b;
  string c = a;
  for (int i = 0; i < n; i++) c[i] = getDist(a[i], b[i]);
  int dist = n;
  stack<int> eq, di;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      eq.push(i);
    } else {
      di.push(i);
    }
  }
  while (dist > t) {
    if (eq.size()) {
      c[eq.top()] = a[eq.top()];
      eq.pop();
      dist--;
    } else if (di.size() >= 2) {
      c[di.top()] = a[di.top()];
      di.pop();
      c[di.top()] = b[di.top()];
      di.pop();
      dist--;
    } else {
      break;
    }
  }
  if (dist > t) {
    cout << -1 << endl;
  } else {
    cout << c << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
char getDist(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 a, b;
  cin >> a >> b;
  string c = a;
  for (int i = 0; i < n; i++) c[i] = getDist(a[i], b[i]);
  int dist = n;
  stack<int> eq, di;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      eq.push(i);
    } else {
      di.push(i);
    }
  }
  while (dist > t) {
    if (eq.size()) {
      c[eq.top()] = a[eq.top()];
      eq.pop();
      dist--;
    } else if (di.size() >= 2) {
      c[di.top()] = a[di.top()];
      di.pop();
      c[di.top()] = b[di.top()];
      di.pop();
      dist--;
    } else {
      break;
    }
  }
  if (dist > t) {
    cout << -1 << endl;
  } else {
    cout << c << endl;
  }
  return 0;
}

```