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>
const int maxn = 100001;
int n, t, e, q;
char a[maxn];
char b[maxn];
char c[maxn];
char next(char a, char b) {
  return a != 'a' && b != 'a' ? 'a' : a != 'b' && b != 'b' ? 'b' : 'c';
}
int main() {
  scanf("%d %d\n", &n, &t);
  e = n - t;
  gets(a);
  gets(b);
  for (int i = 0; i < n; i++) q += a[i] == b[i];
  if (q >= e) {
    for (int i = 0; i < n; i++) {
      if (e && a[i] == b[i]) {
        putchar(a[i]);
        e--;
      } else
        putchar(next(a[i], b[i]));
    }
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (a[i] == b[i]) c[i] = a[i];
  e -= q;
  int ae = e, be = e;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      if (ae) {
        c[i] = a[i];
        ae--;
      } else if (be) {
        c[i] = b[i];
        be--;
      } else {
        c[i] = next(a[i], b[i]);
      }
    }
  }
  puts(!ae && !be ? c : "-1");
  return 0;
}