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;
long long n, t;
string a, b, c, d;
void doc() {
  cin >> n >> t;
  cin >> a;
  a = " " + a;
  cin >> b;
  b = " " + b;
}
char khac(long k) {
  if ('a' != a[k] and b[k] != 'a') return 'a';
  if ('b' != a[k] and b[k] != 'b') return 'b';
  if ('c' != a[k] and b[k] != 'c') return 'c';
}
void xl() {
  c = " ";
  for (int i = 1; i <= n; i++) c += khac(i);
  t = n - t;
  for (int i = 1; i <= n; i++) {
    if (a[i] == b[i] and t > 0) {
      c[i] = a[i];
      t--;
    }
  }
  long na = 0, nb = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] != c[i] and b[i] != c[i] and na < t) {
      c[i] = a[i];
      na++;
    }
    if (a[i] != c[i] and b[i] != c[i] and nb < t) {
      c[i] = b[i];
      nb++;
    }
  }
  if (na < t or nb < t) {
    cout << "-1";
    return;
  }
  for (int i = 1; i <= n; i++) d += c[i];
  cout << d;
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  doc();
  xl();
}