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>
using namespace std;
char s1[100005], s2[100005];
inline char choose(int i) {
  char ch = 'a';
  while (ch == s1[i] || ch == s2[i]) ch++;
  return ch;
}
int main() {
  int n, t, sim = 0;
  scanf("%d %d", &n, &t);
  scanf("%s %s", s1, s2);
  for (int i = 0; i < n; i++) sim += (s1[i] == s2[i]);
  if (n - t <= sim) {
    t = n - t;
    for (int i = 0; i < n; i++) {
      if (t && s1[i] == s2[i]) {
        t--;
        printf("%c", s1[i]);
      } else
        printf("%c", choose(i));
    }
  } else {
    t = n - t;
    t -= sim;
    sim = n - sim;
    if (t * 2 > sim) {
      printf("-1");
      return 0;
    }
    int a = t, b = t;
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i])
        printf("%c", s1[i]);
      else if (a) {
        a--;
        printf("%c", s1[i]);
      } else if (b) {
        b--;
        printf("%c", s2[i]);
      } else
        printf("%c", choose(i));
    }
  }
  return 0;
}