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, s, d, need, av, x, y;
string s1, s2, a, b;
char ans[(int)1e5 + 5];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  cin >> s1 >> s2;
  for (int i = 0; i < n; i++) {
    s += s1[i] == s2[i];
    d += s1[i] != s2[i];
  }
  x = 0;
  need = n - t;
  av = s + d / 2;
  if (need > av) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) continue;
    if (need) {
      ans[i] = s1[i];
      need--;
    } else {
      for (int k = 0; k < 26; k++)
        if (s1[i] - 'a' != k && s2[i] - 'a' != k) {
          ans[i] = 'a' + k;
          break;
        }
    }
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) continue;
    if (need) {
      if (!x) {
        ans[i] = s1[i];
        x = 1 - x;
      } else {
        ans[i] = s2[i];
        x = 1 - x;
        need--;
      }
    } else {
      for (int k = 0; k < 26; k++)
        if (s1[i] - 'a' != k && s2[i] - 'a' != k) {
          ans[i] = 'a' + k;
          break;
        }
    }
  }
  cout << ans << endl;
  return 0;
}