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 n, t, diff = 0;
char s1[100001], s2[100001], s3[100001];
int main() {
  scanf("%d %d\n", &n, &t);
  gets(s1);
  gets(s2);
  for (int i = 0; i < n; i++)
    if (s1[i] != s2[i]) diff++;
  if (diff > t * 2)
    printf("-1\n");
  else {
    if (t >= diff) {
      int cnt = t - diff;
      for (int i = 0; i < n; i++) {
        if (s1[i] != s2[i] || cnt) {
          if (s1[i] == s2[i]) cnt--;
          char c = 'a';
          while (s1[i] == c || s2[i] == c) c++;
          s3[i] = c;
        } else
          s3[i] = s1[i];
      }
    } else {
      int both = t * 2 - diff, each = (t - both) * 2;
      for (int i = 0; i < n; i++) {
        if (s1[i] != s2[i]) {
          if (both > 0) {
            char c = 'a';
            while (s1[i] == c || s2[i] == c) c++;
            s3[i] = c;
            both--;
          } else {
            if (each % 2)
              s3[i] = s1[i];
            else
              s3[i] = s2[i];
            each--;
          }
        } else
          s3[i] = s1[i];
      }
    }
    puts(s3);
  }
  return 0;
}