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, m, k, i, v;
string s, f, ans;
char news(char x, char y) {
  while (true) {
    int z = rand() % 26;
    if (char(z + 'a') != x && char(z + 'a') != y) return char(z + 'a');
  }
}
int main() {
  srand(time(NULL));
  cin >> n >> m;
  cin >> s >> f;
  for (i = 0; i < n; ++i)
    if (s[i] == f[i]) ++k;
  if (m * 2 < (n - k)) {
    cout << -1;
    return 0;
  }
  ans = s;
  k = 0;
  v = m;
  for (i = 0; i < n; ++i) {
    if (m == 0) break;
    if (s[i] != f[i]) m--, ans[i] = f[i];
  }
  for (i = 0; i < n; ++i) {
    if (m == 0) break;
    if (s[i] == f[i]) ans[i] = news(ans[i], f[i]), --m;
  }
  if (m > 0) cout << -1;
  for (i = 0; i < n; ++i)
    if (ans[i] != f[i]) ++k;
  for (i = 0; i < n; ++i) {
    if (k == v) break;
    if (s[i] != f[i]) ans[i] = news(s[i], f[i]), ++k;
  }
  cout << ans;
}