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, ta, tb;
char a[100005], b[100005], res[100005];
char getAnother(char a, char b) {
  for (char c = 'a'; c <= 'z'; c++)
    if (c != a && c != b) return c;
}
int main() {
  cin >> n >> t;
  scanf("%s %s", a, b);
  ta = tb = t;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      if (ta > 0 && ta >= tb) {
        ta--;
        res[i] = b[i];
      } else if (tb > 0) {
        tb--;
        res[i] = a[i];
      } else
        res[i] = 'a';
    } else {
      res[i] = a[i];
    }
  }
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i] && res[i] == a[i] && ta > 0) {
      ta--;
      res[i] = getAnother(a[i], b[i]);
    }
    if (a[i] != b[i] && res[i] == b[i] && tb > 0) {
      tb--;
      res[i] = getAnother(a[i], b[i]);
    }
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && res[i] == a[i] && ta > 0 && tb > 0) {
      ta--;
      tb--;
      res[i] = getAnother(a[i], b[i]);
    }
  }
  ta = tb = 0;
  for (int i = 0; i < n; i++) {
    ta += a[i] != res[i];
    tb += b[i] != res[i];
  }
  if (ta == t && tb == t)
    puts(res);
  else
    puts("-1");
  return 0;
}