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 a[1000000], b[1000000];
int main() {
  cin.tie(nullptr);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  int c1 = n - t;
  int c2 = c1;
  vector<char> s(n);
  char ch;
  for (int i = 0; i < n; i++) s[i] = ' ';
  int i = 0;
  while (i < n) {
    if (s1[i] == s2[i] && c1 > 0 && c2 > 0) {
      s[i] = s1[i];
      c1--;
      c2--;
    }
    i++;
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == ' ' && c1 > 0) {
      s[i] = s1[i];
      c1--;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == ' ' && c2 > 0) {
      s[i] = s2[i];
      c2--;
    }
  }
  if (c1 == 0 && c2 == 0) {
    for (int i = 0; i < n; i++) {
      if (s[i] == ' ') {
        ch = 'a';
        while (s1[i] == ch || s2[i] == ch) ch++;
        s[i] = ch;
      }
    }
    for (int i = 0; i < n; i++) cout << s[i];
  } else
    cout << "-1" << endl;
  return 0;
}