#include <bits/stdc++.h>
using namespace std;
char a[100010];
char b[100010];
char c[100010];
long long flag[100010];
int main() {
  long long t = 1, tc;
  long long i, j, k, l, m, n;
  while (~scanf("%lld", &n)) {
    scanf("%lld", &m);
    scanf("%s", &a);
    scanf("%s", &b);
    memset(flag, -1, sizeof(flag));
    long long sum = 0, sum1 = 0;
    for (i = 0; i < n; i++) {
      if (a[i] != b[i])
        sum++;
      else
        c[i] = a[i];
    }
    sum1 = sum;
    long long tmp = ceil(sum / 2.0);
    if (tmp > m) {
      printf("-1\n");
      continue;
    }
    long long flg = 0, in = -1;
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        if (flg == 0)
          c[i] = a[i];
        else
          c[i] = b[i];
        flag[i] = flg;
        in = i;
        if (flg == 1) m--;
        flg++;
        flg %= 2;
      }
    }
    if (sum % 2 == 1) {
      flag[in] = 2;
      for (j = 'a'; j <= 'z'; j++)
        if (a[in] != j && b[in] != j) {
          c[in] = j;
          break;
        }
      m--;
    }
    for (i = 0; i < n && m > 0; i++) {
      if (a[i] == b[i]) {
        c[i] = a[i] + 1;
        if (c[i] > 'z') c[i] = 'a';
        m--;
      }
    }
    for (i = 0; i < n && m > 0; i++) {
      if (flag[i] == 0 || flag[i] == 1) {
        for (j = 'a'; j <= 'z'; j++)
          if (a[i] != j && b[i] != j) {
            c[i] = j;
            break;
          }
      }
      if (flag[i] == 1) m--;
    }
    c[n] = 0;
    puts(c);
  }
  return 0;
}

### Prompt

Please formulate 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[100010];
char b[100010];
char c[100010];
long long flag[100010];
int main() {
  long long t = 1, tc;
  long long i, j, k, l, m, n;
  while (~scanf("%lld", &n)) {
    scanf("%lld", &m);
    scanf("%s", &a);
    scanf("%s", &b);
    memset(flag, -1, sizeof(flag));
    long long sum = 0, sum1 = 0;
    for (i = 0; i < n; i++) {
      if (a[i] != b[i])
        sum++;
      else
        c[i] = a[i];
    }
    sum1 = sum;
    long long tmp = ceil(sum / 2.0);
    if (tmp > m) {
      printf("-1\n");
      continue;
    }
    long long flg = 0, in = -1;
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        if (flg == 0)
          c[i] = a[i];
        else
          c[i] = b[i];
        flag[i] = flg;
        in = i;
        if (flg == 1) m--;
        flg++;
        flg %= 2;
      }
    }
    if (sum % 2 == 1) {
      flag[in] = 2;
      for (j = 'a'; j <= 'z'; j++)
        if (a[in] != j && b[in] != j) {
          c[in] = j;
          break;
        }
      m--;
    }
    for (i = 0; i < n && m > 0; i++) {
      if (a[i] == b[i]) {
        c[i] = a[i] + 1;
        if (c[i] > 'z') c[i] = 'a';
        m--;
      }
    }
    for (i = 0; i < n && m > 0; i++) {
      if (flag[i] == 0 || flag[i] == 1) {
        for (j = 'a'; j <= 'z'; j++)
          if (a[i] != j && b[i] != j) {
            c[i] = j;
            break;
          }
      }
      if (flag[i] == 1) m--;
    }
    c[n] = 0;
    puts(c);
  }
  return 0;
}

```