#include <bits/stdc++.h>
using namespace std;
int main() {
  int N, T, c = 0;
  char s1[100001], s2[100001], s3[100001];
  scanf("%d%d", &N, &T);
  getchar();
  scanf("%s", s1);
  getchar();
  scanf("%s", s2);
  getchar();
  for (int i = 0; i < N; ++i) {
    if (s1[i] != s2[i]) {
      c++;
    }
  }
  if (c > 2 * T) {
    printf("-1\n");
  } else if (c > T) {
    int i, a = 0;
    for (i = 0; i < N && a < c - T; ++i) {
      if (s1[i] != s2[i]) {
        s3[i] = s1[i];
        a++;
      } else {
        s3[i] = s1[i];
      }
    }
    for (; i < N && a < 2 * (c - T); ++i) {
      if (s1[i] != s2[i]) {
        a++;
      }
      s3[i] = s2[i];
    }
    for (; i < N; ++i) {
      if (s1[i] == s2[i]) {
        s3[i] = s1[i];
      } else {
        if (s1[i] != 'a' && s2[i] != 'a') {
          s3[i] = 'a';
        } else if (s1[i] != 'b' && s2[i] != 'b') {
          s3[i] = 'b';
        } else {
          s3[i] = 'c';
        }
      }
    }
    s3[i] = s1[i];
    printf("%s\n", s3);
  } else {
    int i, a = 0;
    for (i = 0; i < N && a < T - c; ++i) {
      if (s1[i] != 'a' && s2[i] != 'a') {
        s3[i] = 'a';
      } else if (s1[i] != 'b' && s2[i] != 'b') {
        s3[i] = 'b';
      } else {
        s3[i] = 'c';
      }
      if (s1[i] == s2[i]) {
        a++;
      }
    }
    for (; i < N; i++) {
      if (s1[i] == s2[i]) {
        s3[i] = s1[i];
      } else {
        if (s1[i] != 'a' && s2[i] != 'a') {
          s3[i] = 'a';
        } else if (s1[i] != 'b' && s2[i] != 'b') {
          s3[i] = 'b';
        } else {
          s3[i] = 'c';
        }
      }
    }
    s3[i] = s1[i];
    printf("%s\n", s3);
  }
}

### Prompt

Your challenge is to write 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 main() {
  int N, T, c = 0;
  char s1[100001], s2[100001], s3[100001];
  scanf("%d%d", &N, &T);
  getchar();
  scanf("%s", s1);
  getchar();
  scanf("%s", s2);
  getchar();
  for (int i = 0; i < N; ++i) {
    if (s1[i] != s2[i]) {
      c++;
    }
  }
  if (c > 2 * T) {
    printf("-1\n");
  } else if (c > T) {
    int i, a = 0;
    for (i = 0; i < N && a < c - T; ++i) {
      if (s1[i] != s2[i]) {
        s3[i] = s1[i];
        a++;
      } else {
        s3[i] = s1[i];
      }
    }
    for (; i < N && a < 2 * (c - T); ++i) {
      if (s1[i] != s2[i]) {
        a++;
      }
      s3[i] = s2[i];
    }
    for (; i < N; ++i) {
      if (s1[i] == s2[i]) {
        s3[i] = s1[i];
      } else {
        if (s1[i] != 'a' && s2[i] != 'a') {
          s3[i] = 'a';
        } else if (s1[i] != 'b' && s2[i] != 'b') {
          s3[i] = 'b';
        } else {
          s3[i] = 'c';
        }
      }
    }
    s3[i] = s1[i];
    printf("%s\n", s3);
  } else {
    int i, a = 0;
    for (i = 0; i < N && a < T - c; ++i) {
      if (s1[i] != 'a' && s2[i] != 'a') {
        s3[i] = 'a';
      } else if (s1[i] != 'b' && s2[i] != 'b') {
        s3[i] = 'b';
      } else {
        s3[i] = 'c';
      }
      if (s1[i] == s2[i]) {
        a++;
      }
    }
    for (; i < N; i++) {
      if (s1[i] == s2[i]) {
        s3[i] = s1[i];
      } else {
        if (s1[i] != 'a' && s2[i] != 'a') {
          s3[i] = 'a';
        } else if (s1[i] != 'b' && s2[i] != 'b') {
          s3[i] = 'b';
        } else {
          s3[i] = 'c';
        }
      }
    }
    s3[i] = s1[i];
    printf("%s\n", s3);
  }
}

```