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 int N = 1e5 + 3;
int n, t;
string s1, s2;
int a[N];
int X = 0;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> t;
  cin >> s1;
  cin >> s2;
  for (int i = 0; i <= n - 1; i++)
    if (s1[i] == s2[i]) X++;
  if (2 * (n - t) - X > n) {
    cout << -1;
    return 0;
  }
  t = n - t;
  for (int i = 0; i <= n - 1; i++)
    if (s1[i] == s2[i] && t > 0) {
      a[i] = s1[i];
      t--;
    }
  int tmp = t;
  for (int i = 0; i <= n - 1; i++)
    if (a[i] == 0 && s1[i] != s2[i] && t > 0) a[i] = s1[i], t--;
  t = tmp;
  for (int i = 0; i <= n - 1; i++)
    if (a[i] == 0 && s1[i] != s2[i] && t > 0) a[i] = s2[i], t--;
  for (int i = 0; i <= n - 1; i++)
    if (a[i] == 0) {
      for (int j = 'a'; j <= 'z'; j++)
        if (j != s1[i] && j != s2[i]) {
          a[i] = j;
          break;
        }
    }
  for (int i = 0; i <= n - 1; i++) cout << char(a[i]);
}