#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
string a, b, w;
int n, k;
bool us[N];
char dif(char a, char b) {
  for (char e = 'a';; e++) {
    if (e != a && e != b) return e;
  }
}
int main() {
  cin >> n >> k >> a >> b;
  int s = 0, done = 0;
  k = n - k;
  for (int(i) = (0); i < (n); ++(i))
    if (a[i] == b[i] && done < k) {
      done++;
      w += a[i];
      us[i] = 1;
    } else {
      w += dif(a[i], b[i]);
    }
  if (k - done > (n - done) / 2) {
    cout << -1;
    return 0;
  }
  int d1 = k - done;
  int d2 = k - done;
  for (int(i) = (0); i < (n); ++(i))
    if (!us[i]) {
      if (d1) {
        --d1;
        w[i] = a[i];
      } else if (d2) {
        --d2;
        w[i] = b[i];
      } else {
        w[i] = dif(a[i], b[i]);
      }
    }
  cout << w;
  return 0;
}

### Prompt

Generate 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 int N = 100005;
string a, b, w;
int n, k;
bool us[N];
char dif(char a, char b) {
  for (char e = 'a';; e++) {
    if (e != a && e != b) return e;
  }
}
int main() {
  cin >> n >> k >> a >> b;
  int s = 0, done = 0;
  k = n - k;
  for (int(i) = (0); i < (n); ++(i))
    if (a[i] == b[i] && done < k) {
      done++;
      w += a[i];
      us[i] = 1;
    } else {
      w += dif(a[i], b[i]);
    }
  if (k - done > (n - done) / 2) {
    cout << -1;
    return 0;
  }
  int d1 = k - done;
  int d2 = k - done;
  for (int(i) = (0); i < (n); ++(i))
    if (!us[i]) {
      if (d1) {
        --d1;
        w[i] = a[i];
      } else if (d2) {
        --d2;
        w[i] = b[i];
      } else {
        w[i] = dif(a[i], b[i]);
      }
    }
  cout << w;
  return 0;
}

```