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;
char make(char a, char b) {
  for (char i = 'a'; i <= 'z'; i++) {
    if (a != i && b != i) return i;
  }
  return 0;
}
int main() {
  string a, b, c = "";
  int n, t, same = 0, different = 0;
  cin >> n >> t >> a >> b;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i])
      same++;
    else
      different++;
  }
  if (different <= t) {
    t -= different;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        c += make(a[i], b[i]);
      } else if (t > 0 && a[i] == b[i]) {
        c += make(a[i], b[i]);
        t--;
      } else if (a[i] == b[i])
        c += a[i];
    }
  } else if (t != 0 && (different + 1) / 2 <= t) {
    int pairs = different / 2;
    bool first = true;
    for (int i = 0; i < n; i++) {
      if (different != t && a[i] != b[i]) {
        if (first) {
          c += a[i];
          first = false;
        } else {
          c += b[i];
          first = true;
          t--;
          different -= 2;
        }
      } else if (different == t && a[i] != b[i]) {
        c += make(a[i], b[i]);
        different--;
        t--;
      } else if (a[i] == b[i])
        c += a[i];
    }
  } else
    c = "-1";
  cout << c;
  return 0;
}