#include <bits/stdc++.h>
using namespace std;
long n, t;
string a, b;
int main() {
  scanf("%ld%ld", &n, &t);
  cin >> a;
  cin >> b;
  a = '0' + a, b = '0' + b;
  long cnt = 0;
  string res = a;
  for (long i = 1; i <= n; ++i)
    if (a[i] != b[i]) ++cnt;
  if (cnt <= t) {
    for (long i = 1; i <= n; ++i)
      if (a[i] != b[i]) {
        for (long j = 97; j <= 123; ++j)
          if (char(j) != a[i] && char(j) != b[i]) {
            res[i] = char(j);
            break;
          }
      }
    if (cnt == t) {
      res.erase(0, 1);
      cout << res;
      return 0;
    }
    for (long i = 1; i <= n; ++i)
      if (a[i] == b[i]) {
        for (long j = 97; j <= 123; ++j)
          if (char(j) != a[i] && char(j) != b[i]) {
            res[i] = char(j);
            break;
          }
        ++cnt;
        if (cnt == t) break;
      }
    res.erase(0, 1);
    cout << res;
    return 0;
  }
  if (cnt > t) {
    if ((cnt - t) * 2 > cnt) {
      printf("-1");
      return 0;
    }
    long tmp = 0;
    for (long i = 1; i <= n; ++i)
      if (a[i] != b[i]) {
        if (tmp < cnt - t) {
          ++tmp;
          res[i] = a[i];
        } else {
          ++tmp;
          res[i] = b[i];
          if (tmp == 2 * (cnt - t)) {
            for (long j = i + 1; j <= n; ++j)
              if (a[j] != b[j]) {
                for (long k = 97; k <= 123; ++k)
                  if (char(k) != a[j] && char(k) != b[j]) {
                    res[j] = char(k);
                    break;
                  }
              }
            break;
          }
        }
      }
    res.erase(0, 1);
    cout << res;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long n, t;
string a, b;
int main() {
  scanf("%ld%ld", &n, &t);
  cin >> a;
  cin >> b;
  a = '0' + a, b = '0' + b;
  long cnt = 0;
  string res = a;
  for (long i = 1; i <= n; ++i)
    if (a[i] != b[i]) ++cnt;
  if (cnt <= t) {
    for (long i = 1; i <= n; ++i)
      if (a[i] != b[i]) {
        for (long j = 97; j <= 123; ++j)
          if (char(j) != a[i] && char(j) != b[i]) {
            res[i] = char(j);
            break;
          }
      }
    if (cnt == t) {
      res.erase(0, 1);
      cout << res;
      return 0;
    }
    for (long i = 1; i <= n; ++i)
      if (a[i] == b[i]) {
        for (long j = 97; j <= 123; ++j)
          if (char(j) != a[i] && char(j) != b[i]) {
            res[i] = char(j);
            break;
          }
        ++cnt;
        if (cnt == t) break;
      }
    res.erase(0, 1);
    cout << res;
    return 0;
  }
  if (cnt > t) {
    if ((cnt - t) * 2 > cnt) {
      printf("-1");
      return 0;
    }
    long tmp = 0;
    for (long i = 1; i <= n; ++i)
      if (a[i] != b[i]) {
        if (tmp < cnt - t) {
          ++tmp;
          res[i] = a[i];
        } else {
          ++tmp;
          res[i] = b[i];
          if (tmp == 2 * (cnt - t)) {
            for (long j = i + 1; j <= n; ++j)
              if (a[j] != b[j]) {
                for (long k = 97; k <= 123; ++k)
                  if (char(k) != a[j] && char(k) != b[j]) {
                    res[j] = char(k);
                    break;
                  }
              }
            break;
          }
        }
      }
    res.erase(0, 1);
    cout << res;
  }
  return 0;
}

```