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 + 5, MOD = 1e9 + 7;
int i, j, n, k, def, onto = 1;
string a, b;
map<int, char> m;
char c[N];
int main() {
  cin >> n >> k;
  cin >> a >> b;
  for (i = 0; i < n; i++) {
    if (a[i] != b[i]) def++;
  }
  if (def % 2 == 1 && k < def / 2 + 1) return !(cout << "-1");
  if (k < def / 2 || (k == 0 && def)) return !(cout << "-1");
  for (i = 0; i < n; i++) {
    if (a[i] == b[i]) continue;
    if (onto % 2 == 1)
      c[i] = a[i];
    else
      c[i] = b[i];
    onto++;
    if (onto % 2 == 1) k--;
    if (k == 0) break;
  }
  for (i = n - 1; i >= 0; i--) {
    if (k == 0) break;
    if (a[i] == b[i]) continue;
    if (a[i] != 'b' && b[i] != 'b')
      c[i] = 'b';
    else if (a[i] != 'a' && b[i] != 'a')
      c[i] = 'a';
    else
      c[i] = 'c';
    onto++;
    if (onto % 2 == 1) k--;
  }
  i = 0;
  while (k) {
    if (c[i]) {
      i++;
      continue;
    }
    if (a[i] == b[i]) {
      if (a[i] == 'a')
        c[i] = 'b';
      else
        c[i] = 'a';
      k--;
    }
    i++;
  }
  for (; i < n; i++)
    if (a[i] == b[i]) c[i] = a[i];
  for (i = 0; i < n; i++) cout << c[i];
  return 0;
}