#include <bits/stdc++.h>
using namespace std;
char a[100005] = {'\0'}, b[100005] = {'\0'}, ans[100005] = {'\0'};
int main() {
  int n, t, i, temp;
  scanf("%d%d", &n, &t);
  scanf("%s%s", a, b);
  temp = n - t;
  for (i = 0; i < n; i++)
    if (a[i] == b[i] && temp) {
      ans[i] = a[i];
      temp--;
      if (temp <= 0) break;
    }
  if (temp <= 0) {
    for (i = 0; i < n; i++) {
      if (ans[i] == '\0') {
        char tempchar = 'a';
        while (tempchar == a[i] || tempchar == b[i]) tempchar++;
        ans[i] = tempchar;
      }
    }
    for (i = 0; i < n; i++) printf("%c", ans[i]);
    printf("\n");
  } else {
    int tempa, tempb;
    tempa = tempb = temp;
    for (i = 0; i < n; i++)
      if (ans[i] == '\0') {
        ans[i] = a[i];
        tempa--;
        if (tempa <= 0) break;
      }
    for (i += 1; i < n; i++) {
      if (ans[i] == '\0') {
        ans[i] = b[i];
        tempb--;
        if (tempb <= 0) break;
      }
    }
    for (i += 1; i < n; i++) {
      if (ans[i] == '\0') {
        char tempchar = 'a';
        while (tempchar == a[i] || tempchar == b[i]) tempchar++;
        ans[i] = tempchar;
      }
    }
    if (tempa > 0 || tempb > 0)
      printf("-1\n");
    else {
      for (i = 0; i < n; i++) printf("%c", ans[i]);
      printf("\n");
    }
  }
  return 0;
}

### 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;
char a[100005] = {'\0'}, b[100005] = {'\0'}, ans[100005] = {'\0'};
int main() {
  int n, t, i, temp;
  scanf("%d%d", &n, &t);
  scanf("%s%s", a, b);
  temp = n - t;
  for (i = 0; i < n; i++)
    if (a[i] == b[i] && temp) {
      ans[i] = a[i];
      temp--;
      if (temp <= 0) break;
    }
  if (temp <= 0) {
    for (i = 0; i < n; i++) {
      if (ans[i] == '\0') {
        char tempchar = 'a';
        while (tempchar == a[i] || tempchar == b[i]) tempchar++;
        ans[i] = tempchar;
      }
    }
    for (i = 0; i < n; i++) printf("%c", ans[i]);
    printf("\n");
  } else {
    int tempa, tempb;
    tempa = tempb = temp;
    for (i = 0; i < n; i++)
      if (ans[i] == '\0') {
        ans[i] = a[i];
        tempa--;
        if (tempa <= 0) break;
      }
    for (i += 1; i < n; i++) {
      if (ans[i] == '\0') {
        ans[i] = b[i];
        tempb--;
        if (tempb <= 0) break;
      }
    }
    for (i += 1; i < n; i++) {
      if (ans[i] == '\0') {
        char tempchar = 'a';
        while (tempchar == a[i] || tempchar == b[i]) tempchar++;
        ans[i] = tempchar;
      }
    }
    if (tempa > 0 || tempb > 0)
      printf("-1\n");
    else {
      for (i = 0; i < n; i++) printf("%c", ans[i]);
      printf("\n");
    }
  }
  return 0;
}

```