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;
int n, t;
string s1, s2;
int count_matching_indexes(void) {
  int ans = 0;
  for (int i = 0; i < n; ++i)
    if (s1[i] == s2[i]) ++ans;
  return ans;
}
char anything_but(char c, char d) {
  for (char x = 'a'; x <= 'z'; ++x)
    if (x != c and x != d) return x;
  return 0;
}
string solve(void) {
  string ans(n, ' ');
  int pt = n - t;
  int match_ix = count_matching_indexes();
  int rm_1 = max(0, pt - match_ix);
  int rm_2 = rm_1;
  int pmatch = min(match_ix, pt);
  for (int i = 0; i < n; ++i) {
    if (s1[i] != s2[i]) {
      if (rm_1 > 0) {
        ans[i] = s1[i];
        --rm_1;
      } else if (rm_2 > 0) {
        ans[i] = s2[i];
        --rm_2;
      } else
        ans[i] = anything_but(s1[i], s2[i]);
    } else {
      if (pmatch > 0) {
        ans[i] = s1[i];
        --pmatch;
      } else
        ans[i] = anything_but(s1[i], s2[i]);
    }
  }
  if (rm_1 > 0 or rm_2 > 0)
    return "-1";
  else
    return ans;
}
int main(void) {
  cin >> n >> t >> s1 >> s2;
  cout << solve() << endl;
  return 0;
}