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>
char s1[100005], s2[1000005], s3[100005];
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  scanf("%s%s", s1, s2);
  int sam = 0, dif = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i])
      sam++;
    else
      dif++;
  }
  int mostsame = sam + dif / 2;
  if (mostsame < n - t)
    printf("-1\n");
  else {
    int now1 = 0, now2 = 0;
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        if (now1 < n - t) {
          s3[i] = s1[i];
          now1++;
          now2++;
        } else {
          for (s3[i] = 'a'; s3[i] <= 'z'; s3[i]++) {
            if (s3[i] != s1[i]) break;
          }
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        if (now1 == n - t && now2 == n - t) {
          for (s3[i] = 'a'; s3[i] <= 'z'; s3[i]++) {
            if (s3[i] != s1[i] && s3[i] != s2[i]) break;
          }
        } else if (now1 == now2) {
          s3[i] = s1[i];
          now1++;
        } else {
          s3[i] = s2[i];
          now2++;
        }
      }
    }
    s3[n] = 0;
    printf("%s\n", s3);
  }
  return 0;
}