#include <bits/stdc++.h>
using namespace std;
int i, i0, n, m;
char a[100005], b[100005], c[100005];
int main() {
  while (scanf("%d %d %s %s", &n, &m, a, b) != EOF) {
    m = n - m;
    int cnt = 0;
    bool f = 1;
    for (i = 0; i < n; i++)
      if (a[i] != b[i]) cnt++;
    if (cnt % 2 == 0) {
      if (n - cnt / 2 < m)
        f = 0;
      else {
        for (i = 0; i < n; i++) {
          if (a[i] == b[i]) {
            if (m)
              c[i] = a[i], m--;
            else
              c[i] = (a[i] - 'a' + 1) % 26 + 'a';
          }
        }
        m *= 2;
        for (i = 0; i < n; i++) {
          if (a[i] != b[i]) {
            if (m) {
              if (m % 2)
                c[i] = a[i];
              else
                c[i] = b[i];
              m--;
            } else {
              c[i] = 'a';
              while (c[i] == a[i] || c[i] == b[i]) c[i]++;
            }
          }
        }
      }
    } else {
      cnt--;
      int r = 1;
      if (n - cnt / 2 - 1 < m)
        f = 0;
      else {
        for (i = 0; i < n; i++) {
          if (a[i] == b[i]) {
            if (m)
              c[i] = a[i], m--;
            else
              c[i] = (a[i] - 'a' + 1) % 26 + 'a';
          }
        }
        m *= 2;
        for (i = 0; i < n; i++) {
          if (a[i] != b[i]) {
            if (m && r == 0) {
              if (m % 2)
                c[i] = a[i];
              else
                c[i] = b[i];
              m--;
            } else {
              r = 0;
              c[i] = 'a';
              while (c[i] == a[i] || c[i] == b[i]) c[i]++;
            }
          }
        }
      }
    }
    if (f)
      printf("%s\n", c);
    else
      printf("-1\n");
  }
  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;
int i, i0, n, m;
char a[100005], b[100005], c[100005];
int main() {
  while (scanf("%d %d %s %s", &n, &m, a, b) != EOF) {
    m = n - m;
    int cnt = 0;
    bool f = 1;
    for (i = 0; i < n; i++)
      if (a[i] != b[i]) cnt++;
    if (cnt % 2 == 0) {
      if (n - cnt / 2 < m)
        f = 0;
      else {
        for (i = 0; i < n; i++) {
          if (a[i] == b[i]) {
            if (m)
              c[i] = a[i], m--;
            else
              c[i] = (a[i] - 'a' + 1) % 26 + 'a';
          }
        }
        m *= 2;
        for (i = 0; i < n; i++) {
          if (a[i] != b[i]) {
            if (m) {
              if (m % 2)
                c[i] = a[i];
              else
                c[i] = b[i];
              m--;
            } else {
              c[i] = 'a';
              while (c[i] == a[i] || c[i] == b[i]) c[i]++;
            }
          }
        }
      }
    } else {
      cnt--;
      int r = 1;
      if (n - cnt / 2 - 1 < m)
        f = 0;
      else {
        for (i = 0; i < n; i++) {
          if (a[i] == b[i]) {
            if (m)
              c[i] = a[i], m--;
            else
              c[i] = (a[i] - 'a' + 1) % 26 + 'a';
          }
        }
        m *= 2;
        for (i = 0; i < n; i++) {
          if (a[i] != b[i]) {
            if (m && r == 0) {
              if (m % 2)
                c[i] = a[i];
              else
                c[i] = b[i];
              m--;
            } else {
              r = 0;
              c[i] = 'a';
              while (c[i] == a[i] || c[i] == b[i]) c[i]++;
            }
          }
        }
      }
    }
    if (f)
      printf("%s\n", c);
    else
      printf("-1\n");
  }
  return 0;
}

```