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 get_char(char a, char b) {
  for (char i = 'a'; i <= 'z'; i++)
    if (i != a && i != b) return i;
  return 'a';
}
int main() {
  int n, m, i, j, d1, d2;
  string s1, s2, s3;
  cin >> n >> m >> s1 >> s2;
  s3 = string(s1.size(), ' ');
  d1 = n - m;
  d2 = n - m;
  for (i = 0; i < s1.size(); i++) {
    if (!d1 || !d2) break;
    if (s1[i] == s2[i]) {
      s3[i] = s1[i];
      d1--;
      d2--;
    }
  }
  if (d1) {
    for (i = 0; i < s1.size(); i++) {
      if (s3[i] == ' ') {
        s3[i] = s1[i];
        d1--;
      }
      if (!d1) break;
    }
  }
  if (d2) {
    for (i = 0; i < s2.size(); i++) {
      if (s3[i] == ' ') {
        s3[i] = s2[i];
        d2--;
      }
      if (!d2) break;
    }
  }
  for (int i = 0; i < s1.size(); i++) {
    if (s3[i] == ' ') s3[i] = get_char(s1[i], s2[i]);
  }
  if (d1 + d2 == 0)
    cout << s3 << endl;
  else
    cout << -1 << endl;
  return 0;
}