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 main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> t >> s1 >> s2;
  int cur = 0;
  string ans;
  ans.assign(n, '0');
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) cur++;
  }
  if (t < (cur + 1) / 2) {
    cout << -1 << endl;
    exit(0);
  }
  int last = 0;
  for (int i = 0; i < n && t < cur; i++) {
    if (s1[i] != s2[i]) {
      if (last == 1) cur -= 2, t--;
      if (last == 0)
        ans[i] = s2[i];
      else
        ans[i] = s1[i];
      last = 1 - last;
    }
  }
  for (int i = 0; i < n && t > 0; i++) {
    if (ans[i] != '0') continue;
    if (s1[i] != s2[i]) {
      for (int j = 'a'; j <= 'z'; j++) {
        if (s1[i] != j && s2[i] != j) {
          ans[i] = j;
          break;
        }
      }
      t--;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      if (t > 0) {
        if (s1[i] == 'a')
          ans[i] = 'b';
        else
          ans[i] = 'a';
        t--;
      } else
        ans[i] = s1[i];
    }
  }
  cout << ans;
  return 0;
}