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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  char *s1, *s2, *s3, temp;
  long n, t, i, same = 0;
  int flag = 0, prev;
  s1 = (char *)malloc(sizeof(char) * 100000);
  s2 = (char *)malloc(sizeof(char) * 100000);
  s3 = (char *)malloc(sizeof(char) * 100000);
  scanf("%ld %ld", &n, &t);
  scanf("%s %s", s1, s2);
  if (n < t)
    printf("-1\n");
  else {
    t = n - t;
    for (i = 0; i < n; i++) {
      if (same == t) {
        break;
      }
      if (s1[i] == s2[i]) {
        same++;
        s3[i] = s1[i];
      } else if (flag == 1) {
        flag = 0;
        s3[i] = s2[i];
        same++;
      } else if (flag == 0) {
        s3[i] = s1[i];
        flag = 1;
        prev = i;
      }
    }
    if (same != t)
      printf("-1\n");
    else {
      if (flag == 1) {
        if (s1[prev] > 'x') {
          if ((s1[prev] - 1) == s2[prev])
            temp = s1[prev] - 2;
          else
            temp = s1[prev] - 1;
        } else {
          if ((s1[prev] + 1) == s2[prev])
            temp = s1[prev] + 2;
          else
            temp = s1[prev] + 1;
        }
        s3[prev] = temp;
      }
      while (i < n) {
        if (s1[i] > 'x') {
          if ((s1[i] - 1) == s2[i])
            temp = s1[i] - 2;
          else
            temp = s1[i] - 1;
        } else {
          if ((s1[i] + 1) == s2[i])
            temp = s1[i] + 2;
          else
            temp = s1[i] + 1;
        }
        s3[i] = temp;
        i++;
      }
      s3[i] = '\0';
      printf("%s\n", s3);
    }
  }
  return 0;
}