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;
const long long N = 1e6 + 5, K = 501;
string s, t, ans;
long long n, k;
char get(long long i) {
  if (s[i] != 'a' && t[i] != 'a')
    return 'a';
  else if (s[i] != 'b' && t[i] != 'b')
    return 'b';
  else
    return 'c';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long TESTS = 1;
  while (TESTS--) {
    cin >> n >> k;
    cin >> s >> t;
    long long x = 0, y = 0;
    for (long long i = 0; i < n; i++) {
      if (s[i] != t[i]) x++;
    }
    if (k < (x + 1) / 2) {
      cout << -1 << endl;
      return 0;
    }
    for (long long i = 0; i < n; i++) ans += ' ';
    if (k >= x) {
      for (long long i = 0; i < n; i++) {
        if (s[i] != t[i]) ans[i] = get(i);
      }
      k -= x;
      for (long long i = 0; i < n; i++) {
        if (s[i] == t[i]) {
          if (k) {
            ans[i] = get(i);
            k--;
          } else
            ans[i] = s[i];
        }
      }
    } else {
      y = 2 * k - x;
      long long z = 0;
      for (long long i = 0; i < n; i++) {
        if (s[i] != t[i]) {
          if (y) {
            ans[i] = get(i);
            y--;
          } else {
            if (z)
              ans[i] = s[i];
            else
              ans[i] = t[i];
            z ^= 1;
          }
        } else
          ans[i] = s[i];
      }
    }
    cout << ans << endl;
  }
  return 0;
}