#include <bits/stdc++.h>
using namespace std;
const int mx = 1e5 + 3;
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  char s1[mx], s2[mx], s3[mx];
  scanf(" %s %s ", s1, s2);
  int d1 = n, d2 = n;
  for (int i = (0); i < int(n); ++i)
    s3[i] = (s1[i] != 'a' && s2[i] != 'a')   ? 'a'
            : (s1[i] != 'b' && s2[i] != 'b') ? 'b'
                                             : 'c';
  for (int i = (0); i < int(n); ++i) {
    if (d1 == t) break;
    if (s1[i] == s2[i]) {
      s3[i] = s1[i];
      --d1;
      --d2;
    }
  }
  for (int i = (0); i < int(n); ++i) {
    if (s1[i] != s2[i]) {
      if (d1 > t) {
        --d1;
        s3[i] = s1[i];
      } else if (d2 > t) {
        --d2;
        s3[i] = s2[i];
      }
    }
  }
  if (d1 != t || d2 != t) {
    puts("-1");
    return 0;
  }
  for (int i = (0); i < int(n); ++i) printf("%c", s3[i]);
  puts("");
}

### Prompt

In cpp, your task is to solve 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 mx = 1e5 + 3;
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  char s1[mx], s2[mx], s3[mx];
  scanf(" %s %s ", s1, s2);
  int d1 = n, d2 = n;
  for (int i = (0); i < int(n); ++i)
    s3[i] = (s1[i] != 'a' && s2[i] != 'a')   ? 'a'
            : (s1[i] != 'b' && s2[i] != 'b') ? 'b'
                                             : 'c';
  for (int i = (0); i < int(n); ++i) {
    if (d1 == t) break;
    if (s1[i] == s2[i]) {
      s3[i] = s1[i];
      --d1;
      --d2;
    }
  }
  for (int i = (0); i < int(n); ++i) {
    if (s1[i] != s2[i]) {
      if (d1 > t) {
        --d1;
        s3[i] = s1[i];
      } else if (d2 > t) {
        --d2;
        s3[i] = s2[i];
      }
    }
  }
  if (d1 != t || d2 != t) {
    puts("-1");
    return 0;
  }
  for (int i = (0); i < int(n); ++i) printf("%c", s3[i]);
  puts("");
}

```