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 an(char a, char b) {
  if (a != 'a' && b != 'a')
    return 'a';
  else if (a != 'b' && b != 'b')
    return 'b';
  else
    return 'c';
}
int main() {
  double n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  int dif = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) dif++;
  }
  if (t < (dif + 1) / 2 || t > n) {
    cout << -1 << endl;
    return 0;
  }
  string ans;
  for (int i = 0; i < n; i++) ans += '0';
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      ans[i] = an(s1[i], s2[i]);
      t--;
    }
  }
  if (t < 0) {
    int c = 0;
    for (int i = 0; i < n; i++) {
      if (t == 0) break;
      if (s1[i] != s2[i]) {
        if (c % 2 == 0)
          ans[i] = s1[i];
        else
          ans[i] = s2[i];
        t += 0.5;
        c++;
      }
    }
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i]) ans[i] = s1[i];
    }
    cout << ans << endl;
  } else {
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        if (t == 0)
          ans[i] = s1[i];
        else {
          ans[i] = an(s1[i], s2[i]);
          t--;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}