#include <bits/stdc++.h>
using namespace std;
string s1, s2, s3;
char dif(int idx) {
  for (int i = 'a'; i <= 'z'; ++i)
    if (i != s1[idx] && i != s2[idx]) return i;
  return 0;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  s3.resize(n, ' ');
  cin >> s1 >> s2;
  int u = 0;
  vector<int> idx;
  for (int i = 0; i < n; ++i)
    if (s1[i] != s2[i]) u++, idx.push_back(i);
  if (u > 2 * m) {
    puts("-1");
    return 0;
  }
  if (u & 1) {
    s3[idx.back()] = dif(idx.back());
    idx.pop_back();
    u--;
    m--;
  }
  int v = 0;
  while (u && v + u / 2 < m) {
    u -= 2;
    v += 2;
  }
  int rem = m - v - u / 2;
  for (int i = 0; i < v; ++i) s3[idx[i]] = dif(idx[i]);
  for (int i = v; i < (int)idx.size(); i += 2)
    s3[idx[i]] = s1[idx[i]], s3[idx[i + 1]] = s2[idx[i + 1]];
  for (int i = 0; i < n; ++i)
    if (s3[i] == ' ') {
      if (rem)
        s3[i] = dif(i), rem--;
      else
        s3[i] = s1[i];
    }
  cout << s3 << 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;
string s1, s2, s3;
char dif(int idx) {
  for (int i = 'a'; i <= 'z'; ++i)
    if (i != s1[idx] && i != s2[idx]) return i;
  return 0;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  s3.resize(n, ' ');
  cin >> s1 >> s2;
  int u = 0;
  vector<int> idx;
  for (int i = 0; i < n; ++i)
    if (s1[i] != s2[i]) u++, idx.push_back(i);
  if (u > 2 * m) {
    puts("-1");
    return 0;
  }
  if (u & 1) {
    s3[idx.back()] = dif(idx.back());
    idx.pop_back();
    u--;
    m--;
  }
  int v = 0;
  while (u && v + u / 2 < m) {
    u -= 2;
    v += 2;
  }
  int rem = m - v - u / 2;
  for (int i = 0; i < v; ++i) s3[idx[i]] = dif(idx[i]);
  for (int i = v; i < (int)idx.size(); i += 2)
    s3[idx[i]] = s1[idx[i]], s3[idx[i + 1]] = s2[idx[i + 1]];
  for (int i = 0; i < n; ++i)
    if (s3[i] == ' ') {
      if (rem)
        s3[i] = dif(i), rem--;
      else
        s3[i] = s1[i];
    }
  cout << s3 << endl;
  return 0;
}

```